Explanation

(a)
Let $P(n) = 3^{2n} - 1$
At $n = 1,\ P(1) = 8$ which is divisible by 8.
$\therefore P(1)$ is true.
Let $P(k)$ is true, then
$P(k) \equiv 3^{2k} - 1 = 8I$
$\therefore P(k + 1) \equiv 3^{2k + 2} - 1 = (8I + 1)9 - I$
$= 72I + 8 = 8I_{1}$
$\therefore P(n)$ is divisible by 8, $\forall\ n \in N$.