(a) Let $A$ be a commutative ring and $S\subseteq A$ be a multiplicatively closed set. If $S^{-1} A$ is a finite A-module, then $S^{-1} A$ is isomorphic to the A-module $A/\ker i_S$, where $i : A \to S^{-1} A$ is the natural ring homomorphism.
Attempt: I think, by theorem of natural homomorphism on modules, $S^{-1} A $ should always be isomorphic to the $A$-module $A/ \ker i_S$, why is finiteness of $S^{-1}A$ necessary?
(b) Let $A$ be an integral domain , $S_0 = A\setminus\{0\}$ and $K= {S_0}^{-1} A=Q(A)$ be the quotient field of $A$. Then prove that $A=K$ if and only if the canonical homomorphism ${S_0}^{-1} \operatorname{Hom}_A(K,A) \to \operatorname{Hom}_{{S_0}^{-1}A } ({S_0}^{-1} K, {S_0}^{-1} A)$ is surjective.
I am really sorry but for (b), I am not able to make any progress for any side, despite thinking a lot.
Kindly guide!
(a) As mentioned in the comments, if $i_S\colon A\to S^{-1}A$ is surjective, we have the desired isomorphism $A/\ker(i_S)\cong S^{-1}A$.
Let $a_1/s_1,\dots,a_n/s_n\in S^{-1}A$ be a set of generators for the $A$-module $S^{-1}A$. By multiplying appropriate elements of $S$ to the denominators, we may assume $s=s_1=\dots=s_n$. [This is where we use finiteness] Now if $b/t\in S^{-1}A$ is an arbitrary element, then $$b/(st)=\sum_{i=1}^nb_i\cdot a_i/s$$for some $b_i\in A$. Now multiplying $s\in S$ shows $$b/t=\sum_{i=1}^nb_i\cdot a_i/1=i_S\big(\sum_{i=1}^nb_ia_i\big),$$ so $i_S$ is surjective.
(b) Outline: One direction is obvious. Suppose $S_0^{-1}\hom_A(K,A)\to \hom_K(K,K)$ is surjective. In particular, since $\hom_K(K,K)\cong K\ne0$, there must be a nonzero $A$-homomorphism $f\colon K\to A$. This should show that $A=K$.