Let the given statement be $P(n)$, i.e.,

$P(n) : 3^{2n+2}-8n-9$ is divisible by 8.

It can be observed that $P(n) $ is true for $n=1$ Since $ 3^{2 \times 1+2}-8 \times 1-9 = 64$, which is divisible by 8.

Let $P(k)$ be true for some positive integer $k$, i.e.,

$3^{2k+2}-8k-9$ is divisible by 8.

$ \therefore 3^{2k+2}-8k-9 =8m$; where, $m \in N$------(1)

We shall now prove that $P(k+1)$ is true whenever $P(k)$ is true.

Consider

$3^{2(k+1)+2}-8(k+1)-9$

$= 3^{2k+2}.3^2-8k-8-9$

$ = 3^2 (3^{2k+2}-8k-9+8k+9)-8k-17$

$ = 3^2(3^{2k+2}-8k-9)+3^2(8k+9)-8k-17$

$ = 9.8m+9(8k+9)-8k-17$

$ = 9.8m+72k+81-8k-17$

$ = 9.8m+64k+64$

$ = 8(9m+8k+8)$

$ = 8r$ where $r=(9m+8k+8)$ is a natural number.

Therefore, $3^{2(k+1)+2}-8(k+1)-9$ is divisible by 8.

Thus, $P(k+1)$ is true whenever $P(k)$ is true.

Hence, by the principle of mathematical induction, statement $P(n)$ is true for all natural numbers i.e., $n$.