One. (40 points) Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=4, a_{n+1} a_{n}+6 a_{n+1}-4 a_{n}-8=0$. Let $b_{n}=\frac{6}{a_{n}-2}\left(n \in \mathbf{N}_{+}\right)$. Find:
(1) The general term formula of the sequence $\left\{b_{n}\right\}$;
(2) The sum of the first $n$ terms $S_{n}$ of the sequence $\left\{a_{n} b_{n}\right\}$.
(1) From $b_{n}=\frac{6}{a_{n}-2} \Rightarrow a_{n}=\frac{6}{b_{n}}+2$.
Substituting into $a_{n+1} a_{n}+6 a_{n+1}-4 a_{n}-8=0$, we get $b_{n+1}=4 b_{n}+3$.
Then, $b_{n+1}+1=4\left(b_{n}+1\right)$, where $b_{1}+1=4$.
Thus, $b_{n}+1=4 \times 4^{n-1}=4^{n} \Rightarrow b_{n}=4^{n}-1$.
(2) From $b_{n}=\frac{6}{a_{n}-2} \Rightarrow a_{n} b_{n}=2 b_{n}+6$. Therefore,
$$
\begin{array}{l}
S_{n}=a_{1} b_{1}+a_{2} b_{2}+\cdots+a_{n} b_{n} \\
=2\left(b_{1}+b_{2}+\cdots+b_{n}\right)+6 n \\
=2\left(4+4^{2}+\cdots+4^{n}-n\right)+6 n \\
=\frac{8\left(4^{n}-1\right)}{3}-2 n+6 n \\
=\frac{2}{3} \times 4^{n+1}+4 n-\frac{8}{3} .
\end{array}
$$