There are existing results in the literature which answer some of your questions. For instance, I believe that it was known quite early that $\mathrm{Haus}$ is not Cartesian closed. Right now I'd like only to bring to your attention the following result, which dates from 1982.
Theorem (H. Brandenburg and M.Hušek [1]) If a reflective subcategory of $\mathrm{Top}$ contains the two-point discrete space, then it is not Cartesian closed. $\quad\blacksquare$
As the authors remark, the theorem implies that any reflective, Cartesian closed subcategory of $\mathrm{Top}$ must consist of connected spaces.
Corollary $\mathrm{Haus}$ is not Cartesian closed. $\quad\blacksquare$
Later work gives answers in stronger forms. The follow result is contained in papers dating from 1985 and 1990, respectively.
Theorem (Činčura [2], [3]) If a nontrivial reflective subcategory $\mathcal{T}\subseteq\mathrm{Top}$ contains the discrete two-point space and carries a closed symmetric monoidal structure $(\square,[-,-])$, then $\square$ is the $\mathcal{T}$-cross product and $[-,-]$ is the space of continuous maps $X\rightarrow Y$ in the pointwise topology. $\quad\blacksquare$
The pointwise topology is the same as the product topology, so $[X,Y]$ will belong to $\mathcal{T}$ if $\mathcal{T}$ is reflective and $Y\in\mathcal{T}$. The $\mathcal{T}$-cross topology on $X\times Y$ is the initial topology determined by all functions $f\colon X\times Y\rightarrow Z$, where $(i)$ $Z\in\mathcal{T}$, $(ii)$ $f(x,\cdot)\colon Y\rightarrow Z$ is continuous for all $x\in X$, and $(iii)$ $f(\cdot,y)\colon X\rightarrow Z$ is continuous for all $y\in Y$.
The theorem applies to the epireflective subcategory $\mathrm{Haus}$ and clearly implies that it is not Cartesian closed.
Turning now to the various categories of metric spaces, the following is contained in Exercise 6.6.18 of Goubault-Larrecq's textbook [4].
Theorem ([4, Exercise 6.6.18]) The exponential objects in $Met_\infty$ are exactly the discrete metric spaces. $\quad\blacksquare$
If $X$ is a discrete metric space, then for any metric space $Y$, the internal hom $[X,Y]$ is the set of short maps $X\rightarrow Y$ equipped with the sup metric $d(f,g)=\sup_{x\in X}d_Y(f(x),g(x))$. The exercise also includes statements about the categories of extended quasimetric and hemimetric spaces with short maps. It's possible that the methods also work for finite-valued metrics, but I have not verified this.
Corollary $Met_\infty$ is not Cartesian closed. $\quad\blacksquare$
Finally, we let us address the situation for $Met_c$. I had hoped to provide a slightly stronger answer, but didn't quite get there. As such, I apologise for the rather roundabout proof.
Proposition All separable, locally compact metrisable spaces are exponentiable in $Met_c$. Every exponentiable object in $Met_c$ is locally compact.
Since there are metric spaces which are not locally compact ($\mathbb{Q}$ is the simplest example), there are nonexponentiable metric spaces.
Corollary $Met_c$ is not Cartesian closed. $\quad\blacksquare$
The remainder of this post will contain a proof of the statement above. For spaces $Y,Z$ let $C(Y,Z)$ denote the set of continuous functions $Y\rightarrow Z$. Define the $\mathscr{M}$-topology on $C(Y,Z)$ to be the final topology induced by the family of all functions $f\colon X\rightarrow C(Y,Z)$ such that $(i)$ $X$ is metrisable and $(ii)$ the adjoint $f^\flat\colon X\times Y\rightarrow Z$ is continuous.
The $\mathscr{M}$-topology is quite reasonable. It's easily seen to be functorial in both variables. Moreover, it is sequential, as it is the final topology determined by a class of functions with metrisable domains. In fact, if $\mathbb{N}_\infty\cong\{1/n\mid n\in\mathbb{N}\}\cup\{0\}$, then when $Y$ is metrisable the $\mathscr{M}$-topology is exactly the final topology induced by the family of all functions $f\colon \mathbb{N}_\infty\rightarrow C(Y,Z)$ for which the adjoint $f^\flat\colon \mathbb{N}_\infty\times Y\rightarrow Z$ is continuous.
Write $C_k(Y,Z)$ for $C(Y,Z)$ equipped with the compact-open topology.
Lemma For any spaces $Y,Z$, the $\mathscr{M}$-topology on $C(Y,Z)$ contains the compact-open topology. If $Y$ and $C_k(Y,Z)$ are metrisable, then the $\mathscr{M}$-topology coincides with the compact-open topology.
Proof It's well known that for any space $X$, any map $X\times Y\rightarrow Z$ has a continuous adjoint $X\rightarrow C_k(Y,Z)$. By definition, the $\mathscr{M}$-topology is the largest topology enjoying this property for metrisable $X$. Thus it contains the compact-open topology.
Now assume that $Y$ is metrisable. Then for any metrisable space $X$ and any map $f\colon X\rightarrow C_k(Y,Z)$, the adjoint $f^\flat\colon X\times Y\rightarrow Z$ is continuous [5, Corollary 3.2, p.261]. By definition of the $\mathscr{M}$-topology, $f$ is continuous as a map $f\colon X\rightarrow C_\mathscr{M}(Y,Z)$. In case $C_k(Y,Z)$ is metrisable, take $X=C_k(Y,Z)$ and $f=id$ to generate a continuous bijection $C_k(Y,Z)\rightarrow C_\mathscr{M}(Y,Z)$, implying that the compact-open topology contains the $\mathscr{M}$-topology. $\quad\blacksquare$
Lemma Suppose that $Y$ is exponentiable in $Met_c$. Then for any metrisable space $Z$ the exponential $[Y,Z]$ is the set $C(Y,Z)$ equipped with the $\mathscr{M}$-topology.
Proof It's clear that $[Y,Z]$ is the set $C(Y,Z)$ equipped with a metrisable topology. Moreover, for any metrisable space $X$, any map $X\times Y\rightarrow Z$ has a continuous adjoint $X\rightarrow[Y,Z]$. Thus the $\mathscr{M}$-topology contains the exponential topology.
Conversely, notice that the evaluation $[Y,Z]\times Y\rightarrow Z$ is continuous. Since $[Y,Z]$ is metrisable, the adjoint $[Y,Z]\rightarrow C_\mathscr{M}(Y,Z)$ is continuous. This map is clearly bijective, implying that the exponential topology contains the $\mathscr{M}$-topology. $\quad\blacksquare$
Lemma If $Y$ is a k-space and $Z$ is any space, then $C_\mathscr{M}(Y,Z)$ is the sequential coreflection of $C_k(Y,Z)$.
Proof The identity $C_\mathscr{M}(Y,Z)\rightarrow C_k(Y,Z)$ is continuous and, as noted above, $C_\mathscr{M}(Y,Z)$ is sequential. Since $\mathbb{N}_\infty$ is compact metric, $X\times Y$ is a k-space. Thus any map $f\colon\mathbb{N}_\infty\rightarrow C_k(Y,Z)$ has a continuous adjoint $\mathbb{N}_\infty\times Y\rightarrow Z$ by [5, p.261]. Adjointing back yields a continuous map $\mathbb{N}_\infty\rightarrow C_\mathscr{M}(Y,Z)$, since $\mathbb{N}_\infty$ is metrisable. This is exactly a continuous lift of $f$. This shows that $C_\mathscr{M}(Y,Z)$ is the sequential coreflection of $C_k(Y,Z)$. $\quad\blacksquare$
Let $S_\omega$ denote the sequential fan. That is, $S_\omega$ is the quotient of a sum of countably many copies of $\mathbb{N}_\infty$ obtained by identifying all limits points. The space $S_\omega$ is Fréchet-Urysohn, but not metrisable. In particular, any space containing a copy of $S_\omega$ cannot be metrisable.
Theorem (Gruenhage, Tsaban, Zdomskyy [6. Theorem 2.2]) If $Y$ is a metrisable and not locally compact, then $C_k(Y,\mathbb{R})$ contains a closed copy of $S_\omega$. $\quad\blacksquare$
Using this, we have the following.
Proposition If $Y$ is metrisable and not locally compact, then $C_\mathscr{M}(Y,\mathbb{R})$ is not metrisable.
Proof By the theorem above there is a closed embedding $j\colon S_\omega\hookrightarrow C_k(Y,\mathbb{R})$. Since $C_\mathscr{M}(Y,\mathbb{R})$ is the sequential coreflection of $C_k(Y,\mathbb{R})$ and $S_\omega$ is sequential, $j$ lifts to $C_\mathscr{M}(X,\mathbb{R})$ as a closed embedding. Since $S_\omega$ is not metrisable, neither is $C_\mathscr{M}(Y,\mathbb{R})$. $\quad\blacksquare$
In fact, as Gruenhage, Tsaban, and Zdomskyy point out, a topological group contains a closed copy of $S_\omega$ if and only if it contains a closed copy of the Arens space $S_2$. Since the Arens space is not Fréchet-Urysohn, by the same argument as above, $C_\mathscr{M}(Y,\mathbb{R})$ is not Fréchet-Urysohn if $Y$ is metrisable and not locally compact.
Corollary If $Y$ is metrisable, but not locally compact, then $Y$ is not exponentiable in $Met_c$.
Proof If $Y$ is exponentiable, then $[Y,\mathbb{R}]\cong C_\mathscr{M}(Y,\mathbb{R})$. But this contradicts the fact that $C_\mathscr{M}(Y,\mathbb{R})$ is not metrisable. $\quad\blacksquare$
This shows the important half of the main claim. The other part is now shown. Recall that a space $Y$ is hemicompact if it is the union of a family of countably many compact subspace $\{K_n\mid n\in\mathbb{N}\}$ and for any compact subset $K\subseteq Y$ there is $n$ for which $K\subseteq K_n$.
Proposition ([6, Exercise 3.4.E, p.165]) For any metrisable space $Z$, if $Y$ is hemicompact, then $C_k(Y,Z)$ is metrisable. $\quad\blacksquare$
The statement in [6] restricts $Y$ to be Tychonoff, but this is actually not required. Still, if $Y$ is Tychonoff, then $C_k(Y,\mathbb{R})$ is metrisable if and only if $Y$ is hemicompact.
Corollary A separable metrisable space is exponentiable in $Met_c$ if and only if it is locally compact.
Proof Let $Y$ be a separable metrisable space. If $Y$ is not locally compact, then it is not exponentiable in $Met_c$ by the statement above. On the other hand, if $Y$ is locally compact, then it is hemicompact and $C_k(Y,Z)$ is metrisable for any metrisable space $Z$. By the results above $C_k(Y,Z)\cong C_\mathscr{M}(Y,Z)$ and it's clear that putting $[Y,Z]=C_k(Y,Z)$ defines an exponential object in $Met_c$. $\quad\blacksquare$
Remark One might conjecture that the exponential objects in $Met_c$ are exactly the separable, locally compact, metrisable spaces. However, I do not know if a nonseparable, locally compact, metrisable space $Y$ can be exponentiable. The compact-open topology on $C(Y,Z)$ will never be metrisable, but it is conceivable that the finer $\mathcal{M}$-topology could still be metrisable. The space $Y$ decomposes into a sum of uncountably many open, $\sigma$-compact subspaces $Y=\bigsqcup_{i\in\mathcal{I}}Y_i$ and if $Y$ is exponentiable, then for any metrisable spaces $X,Z$ there is a bijection $C(X,[Y,Z])\cong C(X,\prod_{i\in\mathcal{I}}[Y_i,Z])$. If $Z$ has more than one point, then $\prod_{i\in\mathcal{I}}[Y_i,Z]$ is not metrisable. Still, this isn't strong enough to yield a contradiction.
References
[1] H. Brandenburg, M. Hušek , A Remark on Cartesian Closedness, in: Kamps, K.H., Pumplün, D., Tholen, W. (eds) Category Theory. Lecture Notes in Mathematics, vol 962. Springer, Berlin, Heidelberg.
[2] J. Činčura, Closed Structures on Categories of Topological Spaces, Top. and App. 20(2) (1985), 179-189.
[3] J. Činčura, Closed Structures on Reflective Subcategories of the Category of Topological Spaces, Top. and App. 37 (1990), 237-248.
[4] E. Goubault-Larrecq, Non-Hausdorff Topology and Domain Theory: Selected Topics in Point-Set Topology, Cambridge University Press (2013).
[5] J. Dugundji, Topology, Allyn-Bacon (1966).
[6] G. Gruenhage, B. Tsaban, L. Zdomskyy, Sequential properties of function spaces with the compact-open topology, Top. and App. 158(3) (2011), 387-391.
[7] R. Engelking, General Topology, Revised Edition, Heldermann Verlag (1989).
