12. Given a function $f(x)$ defined on $\mathbf{R}$ that satisfies
$$
f(1)=\frac{10}{3} \text {, }
$$

and for any real numbers $x, y$, it always holds that
$$
f(x) f(y)=f(x+y)+f(x-y) .
$$

Let the sequence $\left\{a_{n}\right\}$ satisfy
$$
a_{n}=3 f(n)-f(n-1)\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$
(1) Find the general term formula for the sequence $\left\{a_{n}\right\}$;
(2) Let $b_{n}=\frac{24 a_{n}}{\left(3 a_{n}-8\right)^{2}}\left(n \in \mathbf{Z}_{+}\right)$, and $S_{n}$ be the sum of the first $n$ terms of the sequence $\left\{b_{n}\right\}$. Prove: $S_{n}<1$.
12. (1) In equation (1), let $x=1, y=0$, we get $f(1) f(0)=2 f(1)$.
Also, $f(1)=\frac{10}{3}$, then $f(0)=2$.
In equation (1), let $x=n, y=1$, we get
$$
\begin{array}{l}
f(n) f(1)=f(n+1)+f(n-1) \\
\Rightarrow f(n+1)=\frac{10}{3} f(n)-f(n-1) \\
\Rightarrow a_{n+1}=3 f(n+1)-f(n) \\
\quad=9 f(n)-3 f(n-1) \\
\quad=3(3 f(n)-f(n-1))=3 a_{n} .
\end{array}
$$

Given $a_{1}=3 f(1)-f(0)=8$
$$
\Rightarrow a_{n}=8 \times 3^{n-1} \text {. }
$$
(2) From (1), we know
$$
b_{n}=\frac{24 a_{n}}{\left(3 a_{n}-8\right)^{2}}=\frac{3^{n}}{\left(3^{n}-1\right)^{2}} \text {. }
$$

It is easy to prove: for all $k \in \mathbf{Z}_{\text {+ }}$
$$
3^{k}-1 \geqslant \frac{1}{4}\left(3^{k+1}-1\right) \text {. }
$$

Thus, $b_{k}=\frac{3^{k}}{\left(3^{k}-1\right)^{2}}$
$$
\leqslant \frac{4 \times 3^{k}}{\left(3^{k}-1\right)\left(3^{k+1}-1\right)}=2\left(\frac{1}{3^{k}-1}-\frac{1}{3^{k+1}-1}\right)
$$
$\Rightarrow S_{n}=b_{1}+b_{2}+\cdots+b_{n}$
$$
\leqslant 2 \sum_{k=1}^{n}\left(\frac{1}{3^{k}-1}-\frac{1}{3^{k+1}-1}\right)
$$
$$
\begin{array}{l}
=2\left(\frac{1}{3^{1}-1}-\frac{1}{3^{n+1}-1}\right) \\
=1-\frac{2}{3^{n+1}-1}<1 .
\end{array}
$$