Question: Question: Let $\{\vert\psi\_1\rangle, \vert\psi\_2\rangle \}$ be an orthonormal basis. Define Hamiltonian $\hat{H} = \alpha \left( \vert\psi\_1\rangle \langle\psi\_2\vert + \vert\psi\_2\rangle\langle\psi\_1\vert \right)$ for some real, positive $\alpha$. Find the eigenvalues and eigenkets $\{ \vert E\_1 \rangle, \vert E\_2\rangle \}$ of $\hat{H}$ and express the eigenkets as linear combinations of the $\vert \psi\_k \rangle$ basis states.

I set $\vert E\_1 \rangle = a\_1\vert \psi\_1 \rangle + b\_1\vert\psi\_2\rangle$ and $\vert E\_2 \rangle = a\_2\vert \psi\_1 \rangle + b\_2\vert\psi\_2\rangle$. and

\begin{align}
\hat{H} \vert E\_1 \rangle 
&= \alpha \left(\langle\psi\_2\vert E\_1\rangle\vert\psi\_1\rangle + \langle\psi\_1\vert E\_1\rangle\vert\psi\_2\rangle\right) \\ &= \alpha\left( b\_1\vert\psi\_1\rangle + a\_1\vert\psi\_2\rangle \right).
\end{align}

Because this needs to be in the form of $E\_1\vert E\_1\rangle$, I conclude $a\_1 = b\_1$ and

\begin{equation}
\hat{H}\vert E\_1\rangle = \alpha \vert E\_1\rangle.
\end{equation}

Similarly, for $\vert E\_2\rangle$, I get

\begin{equation}
\hat{H}\vert E\_2\rangle = \alpha \vert E\_2\rangle.
\end{equation}

However, this means the two eigenkets have the same eigenvalue, which does not make sense. I don't know where I did wrong.

Answer: 
Your revised matrix
$$
H=\alpha\left[\matrix{0&1\cr 1&0}\right]
$$
has eigenvalues $\pm \alpha$. You should be able to find the mutually orthonal eigenvectors now.
Whenever you encounter quadratic equations, even without a linear term, there are always two answers. For example, $$x^2=4$$
has two solutions, $x=2$ and $x=-2.$ You have a similar situation in solving for your eigenvalues.

Also, when finding the eigenkets, even if the eigenvalues are the same (which they aren't here), the eigenkets must be constructed to be orthonormal, i.e.,
$$\langle E\_1|E\_2 \rangle = 0 $$