For the first question, the answer is no, an explicit example is given by $A:=\pmatrix{1&0\\ 0&2}$ and $B=\pmatrix{1&1\\ -1&1}$. An other way to see it's not true is the following: take $S$ a symmetric matrix, then you can find $D$ diagonal and $U$ orthogonal (hence unitary) such that $S=U^tDU$ and if $U$ and $D$ commute then $S$ is diagonal.
For the second question, the answer is "not necessarly", because the set $\{a+b\mid a\mbox{ eigenvalue of }A,b\mbox{ eigenvalue of }B\}$ may contain more elements than the dimension of the space we are working with.