12. 已知数列 $\left\{a_{n}\right\}$ 满足 $a_{1}=4, a_{n}=4-\frac{4}{a_{n-1}}(n \geqslant 2)$, 令 $b_{n}=\frac{1}{a_{n}-2}$.

(1)求证: 数列 $\left\{b_{n}\right\}$ 是等差数列;

(2)求数列 $\left\{a_{n}\right\}$ 的通项公式.
(1) $\because a_{n}=4-\frac{4}{a_{n-1}}(n \geqslant 2)$,

$\therefore a_{n+1}=4-\frac{4}{a_{n}}\left(n \in \mathbf{N}^{*}\right)$.

$\therefore b_{n+1}-b_{n}=\frac{1}{a_{n+1}-2}-\frac{1}{a_{n}-2}=\frac{1}{2-\frac{4}{a_{n}}}-\frac{1}{a_{n}-2}=\frac{a_{n}}{2\left(a_{n}-2\right)}-\frac{1}{a_{n}-2}=\frac{a_{n}-2}{2\left(a_{n}-2\right)}=\frac{1}{2}$.

$\therefore b_{n+1}-b_{n}=\frac{1}{2}, n \in \mathbf{N}^{*}$.

$\therefore\left\{b_{n}\right\}$ 是等差数列, 首项为 $\frac{1}{2}$, 公差为 $\frac{1}{2}$.

(2) $b_{1}=\frac{1}{a_{1}-2}=\frac{1}{2}, d=\frac{1}{2}$.

$\therefore b_{n}=b_{1}+(n-1) d=\frac{1}{2}+\frac{1}{2}(n-1)=\frac{n}{2}$.

$\therefore \frac{1}{a_{n}-2}=\frac{n}{2}, \therefore a_{n}=2+\frac{2}{n}$.