Explanation

(a)
$\therefore P\left( \frac{B}{A} \right) = \frac{P(A \cap B)}{P(A)}$
$\Rightarrow \frac{1}{2} = \frac{P(A \cap B)}{\frac{1}{4}}$
$\Rightarrow \ P(A \cap B) = \frac{1}{8}$
Hence, event $A\ and\ B$ are not mutually exclusive.
$\therefore$ Statement 2 is incorrect.
$P\left( \frac{A}{B} \right) = \frac{P(A \cap B)}{P(B)}$
$\Rightarrow P(B) = \frac{P(A \cap B)}{P\left( \frac{A}{B} \right)} = \frac{\frac{1}{8}}{\frac{1}{4}}$
$\Rightarrow P(B) = \frac{1}{2}$
$\therefore\$Event $A\ and\ B$ are independent events.
$P\left( \frac{A^{c}}{B^{c}} \right) = \frac{P(A^{c} \cap B^{c})}{P(B^{c})} = \frac{P\left( A^{c} \right)P(B^{c})}{P(B^{c})}$
$= \frac{3}{4} \bullet \frac{1}{2} \bullet \frac{2}{1} = \frac{3}{4}$
Hence, statement 1 is correct.
Again, $P\left( \frac{A}{B} \right) + P\left( \frac{A}{B^{c}} \right) = \frac{1}{4} + \frac{P(A\ \cap \ B^{c})}{P(B^{c})}$
$= \frac{1}{4} + \frac{P(A) - P(A \cap B)}{P(B^{c})}$
$= \frac{1}{4} + \frac{\frac{1}{4} - \frac{1}{8}}{\frac{1}{2}} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$
Hence, statement 3 is correct.