1) AB = 0 implies that rank(A) + rank (B) $\leq n$ (since range B is in null A). A+B is invertible gives $n =$rank(A+B), which is $\leq$ rank(A) + rank(B) (by dimension of basis). Hence rank(A) + rank(B) = n.
2) [I can only do this over $\mathbb{C}$, need algebraically closed. The result might still hold over $\mathbb{R}$, but I'm not certain due to complex eigenvalues/eigenvectors.] Consider A, B as matrices in $M_n (\mathbb{C})$. AB=BA=0 means that they are simultaneously triangularisable. With this basis, AB=0 means that if a diagonal entry of A is non-zero, then the corresponding entry of B is 0. Likewise, if a diagonal entry of B is non-zero, then the corresponding entry of A is 0. A+B is invertible (in this basis) means that an entry on a diagonal cannot be simultaneously 0. Hence, looking the diagonal entries of A-B are all non-zero, so A-B is invertible. In fact, Det(A-B) = (-1)^k Det(A+B), where k = rank (B).
3) Since rank(M)+null(M) =Dim(M), it follows from 1) that Null(A) + Null(B) = n.