(9) Given the positive sequence $\left\{a_{n}\right\}$ satisfies $\sqrt{a_{n} a_{n+1}+a_{n} a_{n+2}}=4 \sqrt{a_{n} a_{n+1}+a_{n+1}^{2}}+$
$3 \sqrt{a_{n} a_{n+1}}$ and $a_{1}=1, a_{2}=8$, find the general term formula of $\left\{a_{n}\right\}$.
9 By dividing both sides of the known equation by $\sqrt{a_{n} a_{n+1}}$, we get
$$
\sqrt{1+\frac{a_{n+2}}{a_{n+1}}}=4 \sqrt{1+\frac{a_{n+1}}{a_{n}}}+3,
$$

so
$$
\sqrt{1+\frac{a_{n+2}}{a_{n+1}}}+1=4\left(\sqrt{1+\frac{a_{n+1}}{a_{n}}}+1\right) \text {. }
$$

Let $b_{n}=\sqrt{1+\frac{a_{n+1}}{a_{n}}}+1$, then $b_{1}=4, b_{n+1}=4 b_{n}$, which means the sequence $\left\{b_{n}\right\}$ is a geometric sequence with the first term $b_{1}=4$ and common ratio 4, so
$$
b_{n}=b_{1} \cdot 4^{n-1}=4^{n} \text {. }
$$

Therefore, $\sqrt{1+\frac{a_{n+1}}{a_{n}}}+1=4^{n}$, which means
$$
a_{n+1}=\left[\left(4^{n}-1\right)^{2}-1\right] a_{n} .
$$

Thus, when $n>1$,
$$
\begin{aligned}
a_{n} & =\left[\left(4^{n-1}-1\right)^{2}-1\right] a_{n-1} \\
& =\left[\left(4^{n-1}-1\right)^{2}-1\right] \cdot\left[\left(4^{n-2}-1\right)^{2}-1\right] a_{n-2} \\
& =\cdots=\prod_{k=1}^{n-1}\left[\left(4^{k-1}-1\right)^{2}-1\right] a_{1} \\
& =\prod_{k=1}^{n-1}\left[\left(4^{k-1}-1\right)^{2}-1\right],
\end{aligned}
$$

Therefore,
$$
a_{n}=\left\{\begin{array}{ll}
1, & n=1 \\
\prod_{k=1}^{n-1}\left[\left(4^{k-1}-1\right)^{2}-1\right], & n \geqslant 2
\end{array}\right.
$$