I have been stuck on this question for the majority of the day, and I just cant see how part b) is possible.
question is conveyed correctly 
So solving the characteristic polynomial I get $$\hat{H}\:\:E_n\mid E\,\rangle= E_n\mid E\,\rangle$$
Re-arranging and applying the identity matrix $$\hat{H}\:\:E_n\mid E\,\rangle-\:E_n\mid E\,\rangle=0$$
simplifying $$\hat H -IE_n=0$$
$$\begin{pmatrix}\mu &-\mu \\ -\mu &2\mu \end{pmatrix}-E_n\begin{pmatrix}1&0\\ 0&1\end{pmatrix}=0$$
subtracting the matrices and finding the detriment I have the following characteristic equation.
$$\left(\mu -E_n\right)\left(2\mu -E_n\right)-\left(-\mu \right)\left(-\mu \right)=E_{n}^2-3\mu E_n+\mu ^2=0$$
Char Polynomial
$$E_{n}^2-3\mu E_n+\mu ^2=0$$
Solving the polynomial I et the roots
$$E_0=-\frac{1}{2}\left(\sqrt{5}-3\right)\mu \:,\:E_1=\frac{1}{2}\left(\sqrt{5}+3\right)\mu$$
and this is where I am now stumped because how can you b in terms of $$\mu$$ because if I use the roots to find the eigenvector like so:
$$\begin{pmatrix}\mu &-\mu \\ -\mu &2\mu \:\end{pmatrix}\begin{pmatrix}A\\ B\end{pmatrix}=-\frac{1}{2}\left(\sqrt{5}-3\right)\mu \begin{pmatrix}A\\ B\end{pmatrix}$$
$$A\mu -B\mu =-\frac{1}{2}\left(\sqrt{5}-3\right)\mu \:A\:$$
solving for B
I get
$$\frac{-1+\sqrt{5}}{2}A=B\:\:$$
Setting A=1, I get the eigenvector for energy in ground state to be
$$|0> =\begin{pmatrix}1\\ \frac{-1+\sqrt{5}}{2}\end{pmatrix}$$
From this I then set the two eignevector equation equal to each other
$$C\begin{pmatrix}1\\ \frac{-1+\sqrt{5}}{2}\end{pmatrix}=N_0\begin{pmatrix}1\\ \frac{1}{2}-b\end{pmatrix},\:C=N$$
And as C=N due to them being the same normalising constant they cancel out
$$\begin{pmatrix}1\\ \frac{-1+\sqrt{5}}{2}\end{pmatrix}=\begin{pmatrix}1\\ \frac{1}{2}-b\end{pmatrix}$$
But how I make be a function of $$\mu$$ is confusing me because there is no $$\mu$$ in either of these questions.