Since you're only asking for a hint I will point out what I think you're missing:

The integral for the expectation value will yield 4 terms $$\langle x \rangle (t) = \int_{-\infty}^\infty \psi(x,t)^* x \psi(x,t) dx $$ $$= |A|^2 \int_{\infty}^{\infty} \left( -i\sqrt{2}\phi_1^* e^{+iE_1t/\hbar} + \sqrt{3} \phi_2^* e^{+iE_2t/\hbar}\right)x\left(i\sqrt{2}\phi_1e^{-iE_1t/\hbar} + \sqrt{3}\phi_2 e^{-iE_2t/\hbar}\right)dx$$ $$ = |A|^2 \int_{-\infty}^{\infty}\left(2\phi_1^*x\phi_1 + 3\phi_2^*x\phi_2 + i\sqrt{6}\left( \phi_2x\phi_1^*e^{i(E_2-E_1)t/\hbar} - \phi_1^* x \phi_2 e^{-i(E_2-E_1)t/\hbar}\right)\right)dx$$

The expressions $\phi_1^*\phi_1$ and $\phi_2^*\phi_2$ are even functions in position space, when standing alone. Remember that if you integrate over an even function multiplied by an odd function (e.g. $x$) it evaluates to zero:

$$\int_{-\infty}^{\infty} \phi_1^* \phi_1 dx = 1$$ However $$\int_{-\infty}^{\infty} \phi_1^* x \phi_1 dx = 0$$ Hence, you actually lose the first two terms (not the 'cross' terms like you mentioned).

After you remove those two terms you are left with the following expression to evaluate.

$$ = |A|^2 \int_{-\infty}^{\infty}\left(i\sqrt{6}\left( \phi_2x\phi_1^*e^{i(E_2-E_1)t/\hbar} - \phi_1^* x \phi_2 e^{-i(E_2-E_1)t/\hbar}\right)\right)dx$$

And remember that $$ \sin(\theta) = \frac{e^{i\theta} - e^{-i\theta}}{2i}$$

Since you're only asking for a hint I will point out what I think you're missing:

The integral for the expectation value will yield 4 terms $$\langle x \rangle (t) = \int_{-\infty}^\infty \psi(x,t)^* x \psi(x,t) dx $$ $$= |A|^2 \int_{\infty}^{\infty} \left( -i\sqrt{2}\phi_1^* e^{+iE_1t/\hbar} + \sqrt{3} \phi_2^* e^{+iE_2t/\hbar}\right)x\left(i\sqrt{2}\phi_1e^{-iE_1t/\hbar} + \sqrt{3}\phi_2 e^{-iE_2t/\hbar}\right)dx$$ $$ = |A|^2 \int_{-\infty}^{\infty}\left(2\phi_1^*x\phi_1 + 3\phi_2^*x\phi_2 + i\sqrt{6}\left( \phi_2x\phi_1^*e^{i(E_2-E_1)t/\hbar} - \phi_1^* x \phi_2 e^{-i(E_2-E_1)t/\hbar}\right)\right)dx$$

The expressions $\phi_1^*\phi_1$ and $\phi_2^*\phi_2$ are even functions in position space, when standing alone. Remember that if you integrate over an even function multiplied by an odd function (e.g. $x$) it evaluates to zero:

$$\int_{-\infty}^{\infty} \phi_1^* \phi_1 dx = 1$$ However $$\int_{-\infty}^{\infty} \phi_1^* x \phi_1 dx = 0$$ Hence, you actually lose the first two terms (not the 'cross' terms like you mentioned).

On the other hand, for your cross terms, you have $$\int_{-\infty}^{\infty} \phi_1^* \phi_2 dx = 0$$ However, this is your mistake (which Will pointed out in his comment) $$\int_{-\infty}^{\infty} \phi_1^* x \phi_2 dx \neq 0$$ After you remove the appropriate terms you are left with the following expression to evaluate.

$$ = |A|^2 \int_{-\infty}^{\infty}\left(i\sqrt{6}\left( \phi_2x\phi_1^*e^{i(E_2-E_1)t/\hbar} - \phi_1^* x \phi_2 e^{-i(E_2-E_1)t/\hbar}\right)\right)dx$$

And remember that $$ \sin(\theta) = \frac{e^{i\theta} - e^{-i\theta}}{2i}$$