Real matrices with non-real eigenvalues 
I know this covers a lot, so perhaps someone could redirect me to a helpful website. 
for a) I have no idea where to start on the proof, as I don't understand why this is true.
for b) I also have no idea.
for c) A is a real matrix, so if one eigenvalue is complex, the others are too? (the conjugate of the first)?

Hints have already been given for (a), (b), so I'll address (c). There are definitely matrices with both real and non-real eigenvalues: as a trivial example, let $A$ have non-real eigenvalues, and be $n \times n$ and consider the block matrix:
$$B = \begin{pmatrix} k&  0 \\ 0 & A \end{pmatrix}$$
where $k$ is real. This adds an eigenvector of the form $(1,0,...,0)^T$, with real eigenvalue $k$. On the other hand, for all eigenvectors $v$ of $A$, the vector $(0, v^T)^T$ is an eigenvector of $B$ with the same eigenvalue. However, you are right to note that if a real matrix has a non-real eigenvalue, then the conjugate is also an eigenvalue.  This is because the characteriztic polynomial of a real matrix has real coefficients, and real polynomials always have root sets symmetrix under conjugation.
