Membership problem is decidable as it can be solved by parsers.

Finiteness problem is decidable for FSAs (also for CFGs), as we just need to check for a loop in the DFA.

Equivalence problem for FSAs is decidable as we can take the complement of one FSA (complement of FSA is another FSA), and do an intersection with the other (FSAs are closed under intersection also), giving a new FSA. If this new FSA accept no string, then the given FSAs are equivalent, else not equivalent.