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This is true.

Let $P(x) = \sum_{n=1}^{l-1} \frac{x^n}{n} \in {\mathbb F}_l[x]$ (this is $x^{l-1} P_l(x^{-1})$). Then we have $P(1-x) = P(x)$ (check that they have the same derivative and the same value at $1/2$) and $x^l P(x^{-1}) = -P(x)$.

Now if $l \equiv 1 \bmod 3$, then there is a primitive sixth root of unity $\omega \in {\mathbb F}_l$. Note that $1-\omega = \omega^{-1}$. We then obtain $$ \omega P(\omega^{-1}) = \omega^l P(\omega^{-1}) = -P(\omega) $$ and $$ P(\omega^{-1}) = P(1 - \omega) = P(\omega) .$$ So $(1+\omega) P(\omega^{-1}) = 0$, which implies (since $\omega \neq -1$) that $P(\omega) = P(\omega^{-1}) = 0$. These two are actually double roots, since the derivative also vanishes.

In fact, this argument shows that $x^2 - x + 1$ divides $P(x)$ whenever $l > 3$.

Another remark: Since $$ P'(x) = 1 + x + x^2 + \ldots + x^{l-2} = \frac{x^{l-1}-1}{x-1} = \prod_{a \in {\mathbb F}_l \setminus \{0,1\}}(x-a), $$ a root $a \in \overline{\mathbb F}_l$ of $P$ is a multiple root (and then of multiplicity 2) if and only if $a \in {\mathbb F}_l \setminus \{0,1\}$.

(Added later:) One can also give a "combinatorial" proof. The relations satisfied by $P$ imply that the roots of $P$ come in orbits of the $S_3$-action on ${\mathbb P}^1_{\mathbb F_l}$ generated by $a \mapsto 1-a$ and $a \mapsto a^{-1}$ (we think of $P$ has a polynomial of degree $l$ with "leading coefficient" zero, so that $\infty$ is a root). These orbits have size 6, with only three exceptions: $\{0,1,\infty\}$, $\{-1,2,1/2\}$ and $\{\omega, \omega^{-1}\}$ (this uses $l > 3$). We always have the first orbit, which leaves $l-3$ roots. Any orbit other than the third special orbit will contribute a multiple of 3 to the number of roots. So the contribution of the two-element orbit is $\equiv l \bmod 3$, and it can only be 0, 2 or 4. So for $l \equiv 1 \bmod 3$, it must be 4, implying that both are double roots and hence in ${\mathbb F}_l$, and if $l \equiv 2 \bmod 3$, the contribution must be 2, so we have simple roots and they are not in ${\mathbb F}_l$. (This actually proves that $\omega \in {\mathbb F}_l$ iff $l \equiv 1 \bmod 3$ without using that ${\mathbb F}_l^\times$ is cyclic.)

This is true.

Let $P(x) = \sum_{n=1}^{l-1} \frac{x^n}{n} \in {\mathbb F}_l[x]$ (this is $x^{l-1} P_l(x^{-1})$). Then we have $P(1-x) = P(x)$ (check that they have the same derivative and the same value at $1/2$) and $x^l P(x^{-1}) = -P(x)$.

Now if $l \equiv 1 \bmod 3$, then there is a primitive sixth root of unity $\omega \in {\mathbb F}_l$. Note that $1-\omega = \omega^{-1}$. We then obtain $$ \omega P(\omega^{-1}) = \omega^l P(\omega^{-1}) = -P(\omega) $$ and $$ P(\omega^{-1}) = P(1 - \omega) = P(\omega) .$$ So $(1+\omega) P(\omega^{-1}) = 0$, which implies (since $\omega \neq -1$) that $P(\omega) = P(\omega^{-1}) = 0$. These two are actually double roots, since the derivative also vanishes.

In fact, this argument shows that $x^2 - x + 1$ divides $P(x)$ whenever $l > 3$.

Another remark: Since $$ P'(x) = 1 + x + x^2 + \ldots + x^{l-2} = \frac{x^{l-1}-1}{x-1} = \prod_{a \in {\mathbb F}_l \setminus \{0,1\}}(x-a), $$ a root $a \in \overline{\mathbb F}_l$ of $P$ is a multiple root (and then of multiplicity 2) if and only if $a \in {\mathbb F}_l \setminus \{0,1\}$.

(Added later:) One can also give a "combinatorial" proof. The relations satisfied by $P$ imply that the roots of $P$ come in orbits of the $S_3$-action on ${\mathbb P}^1_{\mathbb F_l}$ generated by $a \mapsto 1-a$ and $a \mapsto a^{-1}$ (we think of $P$ has a polynomial of degree $l$ with "leading coefficient" zero, so that $\infty$ is a root). These orbits have size 6, with only three exceptions: $\{0,1,\infty\}$, $\{-1,2,1/2\}$ and $\{\omega, \omega^{-1}\}$ (this uses $l > 3$). We always have the first orbit, which leaves $l-3$ roots. Any orbit other than the third special orbit will contribute a multiple of 3 to the number of roots. So the contribution of the two-element orbit is $\equiv l \bmod 3$, and it can only be 0, 2 or 4. So for $l \equiv 1 \bmod 3$, it must be 4, implying that both are double roots and hence in ${\mathbb F}_l$, and if $l \equiv 2 \bmod 3$, the contribution must be 2, so we have simple roots and they are not in ${\mathbb F}_l$. (This actually proves that $\omega \in {\mathbb F}_l$ iff $l \equiv 1 \bmod 3$ without using that ${\mathbb F}_l^\times$ is cyclic.)

(Added Dec 11, 2014:) As Gjergji Zaimi points out in a comment below, we have that $P(x)$ is (for $l > 2$) the image of $$Q_l(x) = \frac{(x-1)^l - (x^l-1)}{l} \in \mathbb Z[x]$$ in ${\mathbb F}_l[x]$ (use that $\frac{1}{l}\binom{l}{k} = \frac{1}{k} \binom{l-1}{k-1} \equiv (-1)^{k-1}/k \bmod l$). For any prime $l > 3$, $Q_l(\omega) = 0$ (where now $\omega \in \mathbb Q(\sqrt{-3})$), which implies that $P(\bar\omega) = 0$ (where $\bar\omega$ is the image of $\omega$ in $\mathbb F_l$ or $\mathbb F_{l^2}$).

This also explains the relation to FLT. (In fact, the observation that $\omega^l + (\omega^{-1})^l = 1$ gives an $l$-adic first case solution of Fermat's equation when $l \equiv 1 \bmod 3$ gives another proof, assuming the statement on FLT in TZE's question.)

This is true.

Let $P(x) = \sum_{n=1}^{l-1} \frac{x^n}{n} \in {\mathbb F}_l[x]$ (this is $x^l P_l(x^{-1})$). Then we have $P(1-x) = P(x)$ (check that they have the same derivative and the same value at $1/2$) and $x^l P(x^{-1}) = -P(x)$.

Now if $l \equiv 1 \bmod 3$, then there is a primitive sixth root of unity $\omega \in {\mathbb F}_l$. Note that $1-\omega = \omega^{-1}$. We then obtain $$ \omega P(\omega^{-1}) = \omega^l P(\omega^{-1}) = -P(\omega) $$ and $$ P(\omega^{-1}) = P(1 - \omega) = P(\omega) .$$ So $(1+\omega) P(\omega^{-1}) = 0$, which implies (since $\omega \neq -1$) that $P(\omega) = P(\omega^{-1}) = 0$. These two are actually double roots, since the derivative also vanishes.

In fact, this argument shows that $x^2 - x + 1$ divides $P(x)$ whenever $l > 3$.

Another remark: Since $$ P'(x) = 1 + x + x^2 + \ldots + x^{l-2} = \frac{x^{l-1}-1}{x-1} = \prod_{a \in {\mathbb F}_l \setminus \{0,1\}}(x-a), $$ a root $a \in \overline{\mathbb F}_l$ of $P$ is a multiple root (and then of multiplicity 2) if and only if $a \in {\mathbb F}_l \setminus \{0,1\}$.

(Added later:) One can also give a "combinatorial" proof. The relations satisfied by $P$ imply that the roots of $P$ come in orbits of the $S_3$-action on ${\mathbb P}^1_{\mathbb F_l}$ generated by $a \mapsto 1-a$ and $a \mapsto a^{-1}$ (we think of $P$ has a polynomial of degree $l$ with "leading coefficient" zero, so that $\infty$ is a root). These orbits have size 6, with only three exceptions: $\{0,1,\infty\}$, $\{-1,2,1/2\}$ and $\{\omega, \omega^{-1}\}$ (this uses $l > 3$). We always have the first orbit, which leaves $l-3$ roots. Any orbit other than the third special orbit will contribute a multiple of 3 to the number of roots. So the contribution of the two-element orbit is $\equiv l \bmod 3$, and it can only be 0, 2 or 4. So for $l \equiv 1 \bmod 3$, it must be 4, implying that both are double roots and hence in ${\mathbb F}_l$, and if $l \equiv 2 \bmod 3$, the contribution must be 2, so we have simple roots and they are not in ${\mathbb F}_l$. (This actually proves that $\omega \in {\mathbb F}_l$ iff $l \equiv 1 \bmod 3$ without using that ${\mathbb F}_l^\times$ is cyclic.)

This is true.

Let $P(x) = \sum_{n=1}^{l-1} \frac{x^n}{n} \in {\mathbb F}_l[x]$ (this is $x^{l-1} P_l(x^{-1})$). Then we have $P(1-x) = P(x)$ (check that they have the same derivative and the same value at $1/2$) and $x^l P(x^{-1}) = -P(x)$.

Now if $l \equiv 1 \bmod 3$, then there is a primitive sixth root of unity $\omega \in {\mathbb F}_l$. Note that $1-\omega = \omega^{-1}$. We then obtain $$ \omega P(\omega^{-1}) = \omega^l P(\omega^{-1}) = -P(\omega) $$ and $$ P(\omega^{-1}) = P(1 - \omega) = P(\omega) .$$ So $(1+\omega) P(\omega^{-1}) = 0$, which implies (since $\omega \neq -1$) that $P(\omega) = P(\omega^{-1}) = 0$. These two are actually double roots, since the derivative also vanishes.

In fact, this argument shows that $x^2 - x + 1$ divides $P(x)$ whenever $l > 3$.

Another remark: Since $$ P'(x) = 1 + x + x^2 + \ldots + x^{l-2} = \frac{x^{l-1}-1}{x-1} = \prod_{a \in {\mathbb F}_l \setminus \{0,1\}}(x-a), $$ a root $a \in \overline{\mathbb F}_l$ of $P$ is a multiple root (and then of multiplicity 2) if and only if $a \in {\mathbb F}_l \setminus \{0,1\}$.

(Added later:) One can also give a "combinatorial" proof. The relations satisfied by $P$ imply that the roots of $P$ come in orbits of the $S_3$-action on ${\mathbb P}^1_{\mathbb F_l}$ generated by $a \mapsto 1-a$ and $a \mapsto a^{-1}$ (we think of $P$ has a polynomial of degree $l$ with "leading coefficient" zero, so that $\infty$ is a root). These orbits have size 6, with only three exceptions: $\{0,1,\infty\}$, $\{-1,2,1/2\}$ and $\{\omega, \omega^{-1}\}$ (this uses $l > 3$). We always have the first orbit, which leaves $l-3$ roots. Any orbit other than the third special orbit will contribute a multiple of 3 to the number of roots. So the contribution of the two-element orbit is $\equiv l \bmod 3$, and it can only be 0, 2 or 4. So for $l \equiv 1 \bmod 3$, it must be 4, implying that both are double roots and hence in ${\mathbb F}_l$, and if $l \equiv 2 \bmod 3$, the contribution must be 2, so we have simple roots and they are not in ${\mathbb F}_l$. (This actually proves that $\omega \in {\mathbb F}_l$ iff $l \equiv 1 \bmod 3$ without using that ${\mathbb F}_l^\times$ is cyclic.)

This is true.

Let $P(x) = \sum_{n=1}^{l-1} \frac{x^n}{n} \in {\mathbb F}_l[x]$ (this is $x^l P_l(x^{-1})$). Then we have $P(1-x) = P(x)$ (check that they have the same derivative and the same value at $1/2$) and $x^l P(x^{-1}) = -P(x)$.

Now if $l \equiv 1 \bmod 3$, then there is a primitive sixth root of unity $\omega \in {\mathbb F}_l$. Note that $1-\omega = \omega^{-1}$. We then obtain $$ \omega P(\omega^{-1}) = \omega^l P(\omega^{-1}) = -P(\omega) $$ and $$ P(\omega^{-1}) = P(1 - \omega) = P(\omega) .$$ So $(1+\omega) P(\omega^{-1}) = 0$, which implies (since $\omega \neq -1$) that $P(\omega) = P(\omega^{-1}) = 0$. These two are actually double roots, since the derivative also vanishes.

In fact, this argument shows that $x^2 - x + 1$ divides $P(x)$ whenever $l > 3$.

Another remark: Since $$ P'(x) = 1 + x + x^2 + \ldots + x^{l-2} = \frac{x^{l-1}-1}{x-1} = \prod_{a \in {\mathbb F}_l \setminus \{0,1\}}(x-a), $$ a root $a \in \overline{\mathbb F}_l$ of $P$ is a multiple root (and then of multiplicity 2) if and only if $a \in {\mathbb F}_l \setminus \{0,1\}$.

This is true.

Let $P(x) = \sum_{n=1}^{l-1} \frac{x^n}{n} \in {\mathbb F}_l[x]$ (this is $x^l P_l(x^{-1})$). Then we have $P(1-x) = P(x)$ (check that they have the same derivative and the same value at $1/2$) and $x^l P(x^{-1}) = -P(x)$.

Now if $l \equiv 1 \bmod 3$, then there is a primitive sixth root of unity $\omega \in {\mathbb F}_l$. Note that $1-\omega = \omega^{-1}$. We then obtain $$ \omega P(\omega^{-1}) = \omega^l P(\omega^{-1}) = -P(\omega) $$ and $$ P(\omega^{-1}) = P(1 - \omega) = P(\omega) .$$ So $(1+\omega) P(\omega^{-1}) = 0$, which implies (since $\omega \neq -1$) that $P(\omega) = P(\omega^{-1}) = 0$. These two are actually double roots, since the derivative also vanishes.

In fact, this argument shows that $x^2 - x + 1$ divides $P(x)$ whenever $l > 3$.

Another remark: Since $$ P'(x) = 1 + x + x^2 + \ldots + x^{l-2} = \frac{x^{l-1}-1}{x-1} = \prod_{a \in {\mathbb F}_l \setminus \{0,1\}}(x-a), $$ a root $a \in \overline{\mathbb F}_l$ of $P$ is a multiple root (and then of multiplicity 2) if and only if $a \in {\mathbb F}_l \setminus \{0,1\}$.

(Added later:) One can also give a "combinatorial" proof. The relations satisfied by $P$ imply that the roots of $P$ come in orbits of the $S_3$-action on ${\mathbb P}^1_{\mathbb F_l}$ generated by $a \mapsto 1-a$ and $a \mapsto a^{-1}$ (we think of $P$ has a polynomial of degree $l$ with "leading coefficient" zero, so that $\infty$ is a root). These orbits have size 6, with only three exceptions: $\{0,1,\infty\}$, $\{-1,2,1/2\}$ and $\{\omega, \omega^{-1}\}$ (this uses $l > 3$). We always have the first orbit, which leaves $l-3$ roots. Any orbit other than the third special orbit will contribute a multiple of 3 to the number of roots. So the contribution of the two-element orbit is $\equiv l \bmod 3$, and it can only be 0, 2 or 4. So for $l \equiv 1 \bmod 3$, it must be 4, implying that both are double roots and hence in ${\mathbb F}_l$, and if $l \equiv 2 \bmod 3$, the contribution must be 2, so we have simple roots and they are not in ${\mathbb F}_l$. (This actually proves that $\omega \in {\mathbb F}_l$ iff $l \equiv 1 \bmod 3$ without using that ${\mathbb F}_l^\times$ is cyclic.)