12. Given the sequence $\left\{a_{n}\right\}$, the sum of the first $n$ terms is $S_{n}=-n^{2}+9 n+2\left(n \in \mathbf{N}^{*}\right)$.
(1) Find the general term formula of the sequence $\left\{a_{n}\right\}$;
(2) Let $b_{n}=\frac{1}{n\left(12-a_{n}\right)}\left(n \in \mathbf{N}^{*}\right), T_{n}=b_{1}+b_{2}+\cdots+b_{n}$, does there exist a smallest positive integer $m$, such that the inequality $T_{n}<\frac{m}{32}$ holds for all positive integers $n$? If it exists, find the value of $m$; if not, explain the reason.
12. (1) $a_{n}=\left\{\begin{array}{l}10, n=1 . \\ 10-2 n, n \geqslant 2 .\end{array}\right.$
(2) When $n=1$, $b_{1}=\frac{1}{12-a_{1}}=\frac{1}{2}, T_{1}=b_{1}=\frac{1}{2}$;
When $n \geqslant 2$, $b_{n}=\frac{1}{2 n(n+1)}=\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+1}\right)$,
$T_{n}=b_{1}+b_{2}+\cdots+b_{n}=\frac{1}{2}+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{n+1}\right)=\frac{3 n+1}{4 n+4}$;
Since $T_{1}=\frac{3+1}{4+4}=\frac{1}{2}$, $T_{1}$ also satisfies the above formula. Therefore, $T_{n}=\frac{3 n+1}{4 n+4}$. So $T_{n}$ increases monotonically with $n$, so to make the inequality $T_{n}<\frac{m}{32}$ always hold for all positive integers $n$, it is necessary and sufficient that $\lim _{n \rightarrow \infty} T_{n}<\frac{m}{32}$. Also, $\lim _{n \rightarrow \infty} T_{n}=\frac{3}{4}=\frac{24}{32}$, so the smallest positive integer $m=24$.

Note that $a_{n}$ is in piecewise form, so $b_{n}$ is also piecewise, hence when finding $T_{n}=b_{1}+b_{2}+\cdots+b_{n}$, it is still necessary to discuss $n$ in two cases: $n=1$; $n \geqslant 2$.