For 2. let $A$ be the matrix; calculate $A(cv+dw)$ under the assumption that $v$ and $w$ are eigenvectors belonging to different eigenvectors; see whether the result is consistent with $cv+dw$ being an eigenvector.
For 3. write $w=w_1+iw_2$ where $w_1$ and $w_2$ are real, and then see what $Aw=\lambda w$ tells you about $w_1$ and $w_2$.
EDIT: In view of the length of the discussion in the comments, I'll expand on this last part.
We get $$A(w_1+iw_2)=\lambda(w_1+iw_2)$$ which is $$Aw_1+iAw_2=\lambda w_1+i\lambda w_2$$ Now if $a+bi=c+di$ where $a,b,c,d$ are real (real numbers, or vector with real entries, or matrices with real entries), then necessarily $a=c$ and $b=d$ --- that's what equality means in the complex realm. So we deduce $$Aw_1=\lambda w_1,\qquad Aw_2=\lambda w_2$$ So $w_1$ and $w_2$ are in the eigenspace $V_{\lambda}$. We are told $v_1,\dots,v_k$ is a basis for $V_{\lambda}$, so $$w_1=r_1v_1+\cdots+r_kv_k,\qquad w_2=s_1v_1+\cdots+s_kv_k$$ for some real numbers $r_1,\dots,r_k,s_1,\dots,s_k$. Then $$w=w_1+iw_2=c_1v_1+\cdots+c_kv_k{\rm\ with\ }c_j=r_j+is_j,j=1,\dots,k$$
To compute the eigenvalues solve $det \begin{pmatrix}0-\lambda&-1\\2&3-\lambda\end{pmatrix}=0$. You will get $\lambda=1,2$. These are the two eigenvalues. For each of these you need to solve the system of equations given by $\begin{pmatrix}0-\lambda&-1\\2&3-\lambda\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$. The non-zero solutions are the eigenvectors.
For example with $\lambda=1$ we have the system given by $\begin{pmatrix}-1&-1\\2&2\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$ which has a non zero solution $(1,-1)$.
With $\lambda=2$ we have the system given by $\begin{pmatrix}-2&-1\\2&1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$ which has a non zero solution $(1,-2)$.
In this way you have found both eigenvalues and a corresponding eigenvector for both.
Best Answer
Suppose $v_1, v_2$ are vectors to the eigenvalues $\lambda_1, \lambda_2$, let $S=(v_1,v_2)$. Let $A=diag(\lambda_1,\lambda_2)$, you are looking for $X$ such that $XS=SA$, so you can take $X=SAS^{-1}$.