Time-evolution with a time-dependent Hamiltonian 
Consider a quantum mechanical system whose Hilbert space of states is $\mathbb{C}^2$, and has Hamiltonian
  $$\hat{H}= \begin{pmatrix}
E_0e^{t/w_0} & E_1 \\ 
E_1 & E_0e^{t/w_0}
\end{pmatrix}$$
  (a) Describe the time evolution of the system.

I am not exactly sure how exactly to start-off here, and guessing that the time-evolution operator, $U$ is required.
I know that $\hat{U} = e^{- \frac{i}{\hbar} \hat{H} t}$ but not sure how to proceed further.
 A: The identity $\hat{U} = e^{- \frac{i}{\hbar} \hat{H} t}$ holds only for a time-independent hamiltonian, which does not apply here. Instead, the propagator here is given by the time-ordered exponential $\hat{U}(t_2,t_1) = \mathcal T \left[e^{- \frac{i}{\hbar} \int_{t_1}^{t_2} \hat{H}(t) \mathrm d t}\right]$, which is not particularly useful in this situation.
In your situation, you've got very few options left to you other than direct solution of the coupled Schrödinger equations,
\begin{align}
i\hbar \dot a(t) & = E_0e^{t/w_0} a(t) + E_1 b(t), \\ 
i\hbar \dot b(t) & = E_1 a(t) + E_0e^{t/w_0} b(t).
\end{align}
This one is hard to solve but you can make a start by setting $E_1=0$, in which case both $a$ and $b$ obey the differential equation
$$
i\hbar \dot c(t) = E_0e^{t/w_0} c(t),
$$
whose solution is
$$
c(t)= c(0)e^{-i e^{t/w_0}E_0 w_0/\hbar},
$$
so you can ride on that and define $a(t) = \alpha(t) e^{-i e^{t/w_0}E_0 w_0/\hbar}$ and $b(t) = \beta(t) e^{-i e^{t/w_0}E_0 w_0/\hbar}$, for which the Schrödinger equation simplifies to
\begin{align}
i\hbar \dot \alpha(t) & =E_1 \beta(t) , \\ 
i\hbar \dot \beta(t) & = E_1 \alpha(t) ,
\end{align}
whose solutions are
\begin{align}
\alpha(t) & = \alpha(0) \cos(E_1t/\hbar) -i \beta(0) \sin(E_1t/\hbar)\\
\beta(t)  & = -i\alpha(0) \sin(E_1t/\hbar) + \beta(0) \cos(E_1t/\hbar).
\end{align}
Anything beyond that will depend on exactly what you want to do with those solutions.
A: Actually in this specific case you are helped a bit because 
the time dependence is contained in a term proportional to the unit matrix
$$
\hat H = E_0e^{t/\omega_0}\hat I + E_1\left(\begin{array}{cc} 0&1 \\ 1&0 \end{array}\right) \, .
$$
You can then make the time independent change of basis defined by
$$
U=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}
1&1 \\
1&-1\end{array}\right)
$$
that will take $\hat H$ to 
$$
\hat h= U^{-1}\cdot \hat H \cdot U=
\left(\begin{array}{cc}
e^{t/\omega_0}E_0+E_1& 0 \\
0&e^{t/\omega_0}E_0-E_1
\end{array}\right)
$$
The Schrodinger equation for each component in this basis is uncoupled and of the form:
$$
i\hbar \frac{d}{dt}\vert\phi_{\pm}(t)\rangle = 
\left(e^{t/\omega_0}E_0\pm E_1\right)\vert\phi_{\pm}(t)\rangle
$$
with solution 
$$
\vert \phi_{\pm}(t)\rangle=C_\pm e^{-i(\pm E_1 t +e^{t/\omega_0}E_0\omega_0)/\hbar} 
$$
and you can go back to the original basis.  You can verify that this solution is NOT of the form $e^{-i t H/\hbar}$ precisely because, as noted by @EmilioPisanty, $U(t)=e^{-i t H/\hbar}$ is only valid for time-independent Hamiltonians.
