Three. (Full marks 20 points) Given a sequence of positive numbers $\left\{a_{n}\right\}$ satisfying
$$
\begin{array}{l}
\sqrt{a_{n} a_{n+1}+a_{n} a_{n+2}} \\
=3 \sqrt{a_{n} a_{n+1}+a_{n+1}^{2}}+2 \sqrt{a_{n} a_{n+1}},
\end{array}
$$

and $a_{1}=1, a_{2}=3$. Find the general term formula for $\left\{a_{n}\right\}$.
Divide the original equation by $\sqrt[a]{a}, \overline{a_{n}+1}$, we get
$$
\sqrt{1+\frac{a_{n}+2}{a_{n+1}}}:=3 \sqrt{1+\frac{a_{n+1}}{a_{n}}}+2 \text {. }
$$

Let $b_{n}=\sqrt{1+\frac{a_{n}+1}{a_{n}}}+\mathrm{k}$, then $b_{1}=3$.
Transform (1) into
$$
\sqrt{1+\frac{a_{n+2}}{a_{n+1}}}+1=3\left(\sqrt{1+\frac{a_{n+1}}{a_{n}}}+1\right) \text {, }
$$

then $b_{n+1}=3 b_{n}$.
That is, $\left\{b_{n}\right\}$ is a geometric sequence with the first term $b_{1}$ and common ratio 3.
$$
\begin{array}{l}
\therefore b_{n}=b_{1} \cdot 3^{n-1}=3^{n} . \\
\therefore \sqrt{1+\frac{a_{n+1}}{a_{n}}}+1=3^{n} . \\
\therefore a_{n+1}=\left[\left(3^{n}-1\right)^{2}-1\right] a_{n} .
\end{array}
$$

Thus, when $n>1$,
$$
\begin{aligned}
a_{n} & =\left[\left(3^{n-1}-1\right)^{2}-1\right] a_{n-1} \\
& =\left[\left(3^{n-1}-1\right)^{2}-1\right] \cdot\left[\left(3^{n-2}-1\right)^{2}-1\right] a_{n-2} \\
& =\cdots \\
& =\prod_{k=1}^{n-1}\left[\left(3^{k}-1\right)^{2}-1\right] a_{1} \\
& =\prod_{k=1}^{n-1}\left[\left(3^{k}-1\right)^{2}-1\right] .
\end{aligned}
$$

Therefore, $a_{n}=\left\{\begin{array}{l}1,(n=1) \\ \prod_{k=1}^{n-1}\left[\left(3^{k}-1\right)^{2}-1\right] .(n>1)\end{array}\right.$