Intuitively, if $A,B$ are similar matrices, then they represent the same linear transformation, but in different bases. Using this concept, it must be that the eigenvalue structure of two similar matrices must be the same, since the existence of eigenvalues/eigenvectors does not depend on the choice of basis. So, to answer (1), if the eigenvalue structure is different, such as having different multiplicities, then $A,B$ cannot be similar.
To address (2), the answer is no. If there is matching geometric and algebraic multiplicities, then $A,B$ may have different Jordan block structure; for example, $A$ could have three Jordan blocks of size $2,2,2$, and $B$ could have three Jordan blocks of size $3,2,1$. Then the algebraic and geometric multiplicities of both would be, respectively, $6$ and $3$. However, if both $A,B$ are diagonalizable, then $A = PDP^{-1}$ and $B = QDQ^{-1}$, where $D$ is the diagonal matrix, and hence $A = PQ^{-1} BQP^{-1}$, so in this more specific case, they would be similar.