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 $$w=w_1+iw_2=c_1v_1+\cdots+c_kv_k{\rm\ with\ }c_j=r_j+is_j,j=1,\dots,k$$
The answer is yes. Let's assumme that $v_1, v_2$ are the eigenvectors that correspond to the same eigenvalue $\alpha_1$. Observe that $\lambda_1 v_1 + \lambda_2 v_2$ is an eigenvector for the eigenvalues $\alpha_1$ and is therefore linearly independent of our third vector $v_3$. This means that if $\lambda_1 v_1 + \lambda_2 v_2 + \lambda_3 v_3 = 0$ we necessarily have $\lambda_3 = 0$. Now this implies $\lambda_1 v_1 + \lambda_2 v_2 = 0$, which by assumption yields $\lambda_1 = \lambda_2 = 0$.
Note that this is nothing else than observing that the sum of eigenspaces to different eigenvalues is a direct sum.
Best Answer
All nonzero vectors are eigenvectors, since all vectors $v$ satisfy $Mv=0v$ if $M$ is the zero matrix.