(A)M can either accept, go into loop or halt within K steps(which are finite). So within K steps, we are guaranteed to get an answer whether a string will be accepted or not. So this is Decidable.
(B)Equivalence of two Turing machines would require for any two machines M1 and M2 we must be able to check
L(M1) = L(M2).
Even if two machines halt on every input that is given to them, we cannot keep checking this indefinitely over ∑*.
So, this is undecidable.
(C)This is decidable. Given a finite state machine we can look for a path from initial state to final state. If such a path exists, we can say language accepted by a given FA is not empty.
(D)Decidable : Given a Context-Free grammar G, we can look for useless symbols in grammar. If all the symbols of a production of grammar are useless, surely the language generated by the Context-free Grammar is empty, otherwise not.