Intuitively recognise eigenvalues? How can I recognise eigenvalues without calculation?

 A: The following is a relatively short answer to each of the questions. Be sure to check whether everything is clear; for me these answers would be sufficient, but your teacher might want a more worked out solution, which you should be able to provide.
a) The vectors which lie in the plane you're projecting on, are send to themselves. Thus, there is a two-dimensional eigenspace with eigenvalue $1$, consisting of the vectors in the given plane. On the other hand, the vector $(3, 1, -2)$, which is orthogonal to the plane, is send to $0$ under the given projection (do you see why?). Hence, it is a vector with eigenvalue $0$. Since we can find a two-dimensional eigenspace with eigenvalue $1$ and a third eigenvector with eigenvalue $0$, these are the only eigenvectors (do you see why there can't be more?)
b) If $A$ is an eigenvector with eigenvalue $\lambda$, then
$$
A = \lambda A^T = \lambda^2 (A^T)^T = \lambda^2 A 
$$
such that $\lambda^2 = 1$ and thus $\lambda = 1$ or $\lambda = -1$. (This simple approach was pointed out to me by Semiclassical.) In the first case, the eigenspace consists exactly of the three-dimensional subspace of symmetric matrices. Find an eigenvector corresponding to $\lambda = -1$ yourself. 
c) Since $2$ is always a root of $R(p)$, $p$ can only be an eigenvector if $2$ is a root of $p$ or $\lambda = 0$. Check yourself that there is a one-dimensional eigenspace belonging to $\lambda = 0$. If $\lambda \neq 0$, then an eigenvector $p$ must be of the form $(x-2)(ax + b)$. One can then easily calculate that this is an eigenvector if and only if $b = -2a$. (do this!)
