Explanation

(a)
Given, $P(A \cup B) = 0.6,\ P(A \cap B) = 0.2$
Probability of exactly one of the event occurs is
$P\left( \overline{A} \cap B \right) + P(A \cap \overline{B})$
$= P(B) - P(A \cap B) + P(A) - P(A \cap B)$
$= P(A \cup B) + P(A \cap B) - 2P(A \cap B)$
[$\because P(A \cup B) = P(A) + P(B) - P(A \cap B)\rbrack$
$= P(A \cup B) - P(A \cap B)$
$= 0.6 - 0.2 = 0.4$