X and Y be finite sets and f: X->Y be a function. 
The option D is the correct option. But, I have a doubt since the inverse of function can exist or cannot exist, how can this option be true. How to approach these questions? Should we assume something or the question should be solved in general. Why are other options discarded. Option B may be correct.
 A: If you have a function $f: X \to Y$ between sets, then $f^{-1}$ is usually defined between sets of subsets of the two sets. So for $S\subseteq Y$
$$
f^{-1}(S) = \{x\in X : f(x) \in S\}.
$$
So there is no problem with $f^{-1}$ not being defined. If $S$ and $T$ are subsets of $Y$, then indeed
$$
f^{-1}(S\cap T) = f^{-1}(S) \cap f^{-1}(T).
$$
For example, if $x\in f^{-1}(S\cap T)$, then by definition $f(x) \in S\cap T$. So $f(x) \in S$ and $f(x)\in T$. So $x\in f^{-1}(S)$ and $x\in f^{-1}(T)$. That is $f^{-1}(S) \cap f^{-1}(T).$. This proves that $f^{-1}(S\cap T) \subseteq f^{-1}(S) \cap f^{-1}(T)$.
A similar proof exists for the other way.

Consider the example $f: \mathbb{R} \to \mathbb{R}$ given by $f(x) = x^2$. This function is not injective. So we usually say that the inverse of $f$ doesn't exist. What we mean is that there is no function $g$ such that $f\circ g$ and $g\circ f$ is the identity function. But it is still true, for example, that 
$$
f^{-1}(\{1\}) = \{-1, 1\}.
$$
Sometimes we write $f^{-1}(1)$ instead of $f^{-1}(\{1\})$.
A: Inverse will exist. Whether it will be a function or not is a different question. I hope this helps.
You can find counterexamples for first 3 options.
Like for option B,
X = Y = { 1,2,3,4,5 }
f(1) = 1; f(2) = 2; f(3) = 1; f(4) = 2; f(5) = 5;  
A = { 1,2,3 } ; B = { 3,4,5 }
f(A) = { 1,2 }
f(B) = { 1,2,5 }
f(A^B) = f({ 3 }) = { 1 }
f(A) ^ f(B) = { 1,2 } 
A: Just go through them systematically.A: False let A=B={1}
B: False let X={1,2} Y={1} f(1)=1 f(1)=2 and let A={1} B={2}
C: False a={1} b= {2,3}
D: True by elimination, a sketch proof is provided by Thomas
A: f: X -> Y be a function from X to Y and S be a
subset of Y. We define the inverse image of S to be the subset
of X whose elements are precisely all pre-images of all elements of S. We denote the inverse image of S by 
$f^{-1}(S) = \{x\in X : f(x) \in S\}$
