(1) (40 points) The sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=4, a_{n+1} a_{n}+6 a_{n+1}-4 a_{n}-8=0$, and let $b_{n}=\frac{6}{a_{n}-2}, n \in \mathbf{N}^{*}$.
(1) Find the general term formula for the sequence $\left\{b_{n}\right\}$;
(2) Find the sum of the first $n$ terms $S_{n}$ of the sequence $\left\{a_{n} \cdot b_{n}\right\}$.
1 (1) From $b_{n}=\frac{6}{a_{n}-2}$, we get $a_{n}=\frac{6}{b_{n}}+2$. Substituting into
$$
a_{n+1} a_{n}+6 a_{n+1}-4 a_{n}-8=0,
$$

we get $b_{n+1}=4 b_{n}+3$. Therefore,
$$
b_{n+1}+1=4\left(b_{n}+1\right),
$$

where $b_{1}+1=4$. So
$$
b_{n}+1=4 \cdot 4^{n-1}=4^{n},
$$

hence $b_{n}=4^{n}-1$.
(2) From $b_{n}=\frac{6}{a_{n}-2}$, we get $a_{n} \cdot b_{n}=2 b_{n}+6$, so
$$
\begin{aligned}
S_{n} & =a_{1} b_{1}+a_{2} b_{2}+\cdots+a_{n} b_{n} \\
& =2\left(b_{1}+b_{2}+\cdots+b_{n}\right)+6 n \\
& =2\left(4+4^{2}+\cdots+4^{n}-n\right)+6 n \\
& =\frac{8\left(4^{n}-1\right)}{3}-2 n+6 n \\
& =\frac{2}{3} \cdot 4^{n+1}+4 n-\frac{8}{3} .
\end{aligned}
$$