If $A = \{x \mid x \in N, x \text{ is a prime number less than } 12\}$ and $B = \{x \mid x \in N, x \text{ is a factor of } 10\},$ then $A \cap B = \dots$

Consider the following relations:
$(1) \, A - B = A - (A \cap B)$
$(2) \, A = (A \cap B) \cup (A - B)$
$(3) \, A - (B \cup C) = (A - B) \cup (A - C)$
Which of these is/are correct?

Let $E, F$ and $G$ be three events having probabilities $P(E) = \frac{1}{8}, P(F) = \frac{1}{6}$ and $P(G) = \frac{1}{4}$,and let $P(E \cap F \cap G) = \frac{1}{10}$. For any event $H$,if $H^C$ denotes its complement,then which of the following statements is(are) $TRUE$?
$(A) P(E \cap F \cap G^C) \leq \frac{1}{40}$
$(B) P(E^C \cap F \cap G) \leq \frac{1}{15}$
$(C) P(E \cup F \cup G) \leq \frac{13}{24}$
$(D) P(E^C \cap F^C \cap G^C) \leq \frac{5}{12}$

For any two sets $A$ and $B$, $A-(A-B)$ equals

If $A=\{3, 5, 7, 9, 11\}, B=\{7, 9, 11, 13\}, C=\{11, 13, 15\}$ and $D=\{15, 17\}$,find $A \cap (B \cup C)$.

If $A = \{1, 2, 3, 4\}$,$B = \{3, 4, 5, 6\}$,$C = \{5, 6, 7, 8\}$,and $D = \{7, 8, 9, 10\}$,find $A \cup B$.