I am not undertanding what I have to do in letter (c) of this problem.. I just wrote $\alpha_1=\begin{bmatrix} \cos\phi & -\sin\phi\\ \sin\phi & \cos\phi \end{bmatrix} \begin{bmatrix}1\\0\end{bmatrix}=\begin{bmatrix}\cos\phi \\ \sin\phi\end{bmatrix}$
and $\alpha_2=\begin{bmatrix} \cos\phi & -\sin\phi\\ \sin\phi & \cos\phi \end{bmatrix} \begin{bmatrix}0\\1\end{bmatrix}=\begin{bmatrix}-\sin\phi \\ \cos\phi\end{bmatrix}$
Let $e_1 , e_2$ be standart basis vectors for $R^2$
$\epsilon_1=n_1e_1+n_2e_2$ ,$\epsilon_2=m_1e_1+m_2e_2$
$U_\phi(e_1)=\begin{pmatrix} \cos\phi \\ \sin\phi\end{pmatrix}$ , $U_\phi(e_2)=\begin{pmatrix} -\sin\phi \\ \cos\phi \end{pmatrix}$
Let $v=c_1\alpha_1+c_2\alpha_2$ then
$U_\theta(v)=U_\theta c_1\alpha_1+U_\theta c_2\alpha_2$
$=c_1U_\theta\alpha_1+c_2U_\theta\alpha_2=c_1U_\theta U_\phi\epsilon_1+c_2U_\theta U_\phi\epsilon_2$$=U_\theta U_\phi(c_1\epsilon_1+c_2\epsilon_2)$
$=U_\theta U_\phi(c_1n_1e_1+c_1n_2e_2+c_2m_1e_1+c_2m_2e_2)=(c_1n_1+c_2m_1)U_\theta U_\phi e_1+(c_1n_2+c_2m_2)U_\theta U_\phi e_2$
$=(c_1n_1+c_2m_1)\begin{pmatrix} \cos\phi+\theta \\ \sin\phi+\theta\end{pmatrix}+(c_1n_2+c_2m_2)\begin{pmatrix} -\sin\phi+\theta \\ \cos\phi+\theta \end{pmatrix}$
$\begin{pmatrix} (c_1n_1+c_2m_1)\\ (c_1n_2+c_2m_2) \end{pmatrix}=\begin{pmatrix} n_1 & m_1 \\ n_1 & m_2\end{pmatrix} \begin{pmatrix} c_1 \\ c_2\end{pmatrix}$
so we can say that
$(c_1n_1+c_2m_1)\begin{pmatrix} \cos\phi+\theta \\ \sin\phi+\theta\end{pmatrix}+(c_1n_2+c_2m_2)\begin{pmatrix} -\sin\phi+\theta \\ \cos\phi+\theta \end{pmatrix}=\begin{pmatrix} \cos\phi+\theta & -\sin\theta+\phi \\ \sin \phi+\theta & \cos \phi + \theta \end{pmatrix}\begin{pmatrix} n_1 & m_1 \\ n_1 & m_2\end{pmatrix} \begin{pmatrix} c_1 \\ c_2\end{pmatrix}$
so our matrix is $\begin{pmatrix} \cos\phi+\theta & -\sin\theta+\phi \\ \sin \phi+\theta & \cos \phi + \theta \end{pmatrix}\begin{pmatrix} n_1 & m_1 \\ n_1 & m_2\end{pmatrix} $