True. Let x\in f^{-1}(A\cup B). Then f(x)\in A or f(x)\in B. Therefore x\in f^{-1}(A)\cup f^{-1}(B). Next, let x\in f^{-1}(A)\cup f^{-1}(B). Then f(x)\in A or f(x)\in B. So f(x)\in A\cup B and x\in f^{-1}(A\cup B).

First trial. False. There is a mapping f:X\rightarrow Y of X into Y. Suppose that there is f^{-1}(A)\cap f^{-1}(B) = \phi. And let y \in A \cap B such that A \subset Y and B \subset Y. Then f(a) = y for some a \in f^{-1}(A) and f(b) = y for some b \in f^{-1}(B). For the condition, a and b cannot be in f^{-1}(A)\cap f^{-1}(B).

틀렸댄다. 고민 좀 더 해보자. 흑흑...

Second trial. True. Let x \in f^{-1}(A\cap B). Then f(x)\in A\cap B. This means that x\in f^{-1}(A) and x\in f^{-1}(B). Therefore x\in f^{-1}(A)\cap f^{-1}(B). Next, let x \in f^{-1}(A)\cap f^{-1}(B). Then f(x)\in A and f(x)\in B. That means f(x)\in A\cap B so x\in f^{-1}(A\cap B).

참... 그랬어요

First trial. True. Let y_1\in f(A\cup B), y_2\in f(A)\cup f(B). Then f^{-1}(y_1)\subset A\cup B and f^{-1}(y_2)\subset A or f^{-1}(y_2)\subset B, respectively. We know that what a element is in A\cup B and what a element is in A or B are exactly same.

Second trial. True. Let y\in f(A\cup B). Then f(x)=y for some x\in A\cup B. This means x\in A or x\in B. Therefore y\in f(A)\cup f(B). And let y\in f(A)\cup f(B). Then f(x)=y for some x\in A or x\in B. This means that x\in A\cup B so y\in f(A\cup B).

It is not always true. Suppose that there exists an element y\in f(A)\cap f(B) and A\cap B=\phi. Then there is no element in f(A\cap B).