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
$$ \sin(\theta) = \frac{e^{i\theta} - e^{-i\theta}}{2i}$$
If you represent the wave function $\psi(x)$ with it's fourier transform,
\begin{eqnarray*}
\psi(x) &=& \frac{1}{\sqrt{2\pi \hbar}}\int \tilde{\psi}(p)e^{\frac{ipx}{\hbar}}dp\\
\psi(x)^\star &=& \frac{1}{\sqrt{2\pi \hbar}} \int \tilde{\psi}^\star(q)e^{\frac{-iqx}{\hbar}}dq
\end{eqnarray*}
(where p and q are almost like "dummy" momenta), then you can rewrite the expectation value of momentum as follows:
\begin{eqnarray}
\langle p \rangle &=& \int \psi^\star \left(-i\hbar \frac{\partial}{\partial x}\right)\psi dx\\
&=& \frac{1}{2\pi \hbar} \int \tilde{\psi}^\star(q)e^{\frac{-iqx}{\hbar}}\left(-i\hbar \frac{\partial}{\partial x}\right) \tilde{\psi}^\star(p)e^{\frac{ipx}{\hbar}} dpdqdx
\end{eqnarray}
Now if you apply the derivative with respect to $x$, you'll spit out a $p$ in the integrand
\begin{eqnarray}
&=& \frac{1}{2\pi \hbar} \int \tilde{\psi}^\star(p) \tilde{\psi}^\star(q)e^{\frac{i(q-p)x}{\hbar}} \left(p\right) dpdqdx
\end{eqnarray}
and exchanging integration order to integrate over $x$ first -- since we know these functions to be $L^2$ integrable --yields the (scaled) dirac delta function:
\begin{eqnarray}
&=& \frac{1}{2\pi \hbar} \int \tilde{\psi}^\star(p) \tilde{\psi}^\star(q)\hbar \delta(q-p) \left(p\right) dpdq \\
\langle p \rangle &=&\frac{1}{2\pi } \int \vert\tilde{\psi}(p)\vert^2 p dp
\end{eqnarray}
There's a missing factor of $2\pi$ in there, but I trust you'll find it if you do it carefully by hand.
Best Answer
$\newcommand{\d}{\;\mathrm{d}}$
Elaborating on the answer of zeldredge, I want to say why the following expression works:
$$P(E_0) = \left| \int_{-\infty}^{\infty} \ \Phi^* \psi \d x \right|^2 \tag{1}$$
Notice that the eigenfunctions of the Hamiltonian spans the space. That is you can write any function $\psi$ as a linear combination of eigenfunctions of the Hamiltonian ie.
$$\psi=a_0\Phi_0+a_1\Phi_1+a_2\Phi_2 \dots = \sum_i a_i \Phi_i \tag{2}$$
with $a_i \in \mathbb C$ and $\sum_i \left|a_i \right|^2=1$. Notice that as in all superposition of states the probability of finding particle in a particular state is given by $\left| a_i \right|^2$. Assuming that you have chosen your states cleverly such that they are orthonormal ie they satisfy the following property,
$$\int \Phi_i^*\Phi_j \d x = \delta_{ij}$$
you can see immediately why Eqn. (1) works.
$$\int \Phi_i^* \psi \d x = \int \Phi_i^* \sum_j a_j \Phi_j \d x = \sum_j a_j\int \Phi_i^* \Phi_j \d x = \sum_j \delta_{ij} a_j =a_i $$
Therefore the norm squared of this expression just gives the probability to find the state in the energy eigenfunction $\Phi_i$.
Notice I assumed that the energy levels are discrete to write the equation (2). Furthermore I dropped $x$'s in $\psi(x)$ and $\Phi(x)$ because they were cluttering the equations.