11. Given that the graph of the function $y=f(x)$ is symmetric to the graph of $y=2-\frac{1}{x}$ with respect to the line $y=x$. The sequence $\left\{a_{n}\right\}$ satisfies $a_{n+1} = f\left(a_{n}\right)\left(n \in \mathbf{N}^{*}\right)$.
(1) If $a_{1}=3$, prove that there exists a positive integer $n_{0}$ such that when $n \geqslant n_{0}$, we have $a_{n+1}>a_{n}$;
(2) Let $1+\frac{1}{m}<a_{1}<\frac{m}{m-1}$, where $m$ is a constant, $m \in \mathbf{N}^{*}$ and $m \neq 1$. Find the smallest positive integer $N$ such that when $n \geqslant N$, we always have $0<a_{n}<1$.
11. Solution: (1) Since the graph of the function $y=f(x)$ is symmetric to the graph of $y=2-\frac{1}{x}$ about the line $y=x$, for any point $(x, y)$ on the graph of the function $y=f(x)$, the point $(y, x)$ is on the graph of the function $y=2-\frac{1}{x}$, i.e., $x=2-\frac{1}{y}$, so, $y=f(x)=\frac{1}{2-x}, a_{n+1}=$ $\frac{1}{2-a_{n}}$.
Since when $a_{1}=3$, $a_{2}=-10 \Leftrightarrow a_{n}0$. That is, $a_{n+1}>a_{n}$.
(2) From $a_{n+1}=\frac{1}{2-a_{n}}$ we get $a_{n+1}-1=\frac{1}{2-a_{n}}-1=\frac{a_{n}-1}{2-a_{n}}$, taking the reciprocal gives:
$\frac{1}{a_{n+1}-1}=\frac{2-a_{n}}{a_{n}-1}=\frac{1}{a_{n}-1}-1$, so $\left\{\frac{1}{a_{n}-1}\right\}$ is an arithmetic sequence,
$\frac{1}{a_{n}-1}=\frac{1}{a_{1}-1}-(n-1)=\frac{a_{1}-n\left(a_{1}-1\right)}{a_{1}-1}$, thus $a_{n}=1+\frac{a_{1}-1}{a_{1}-n\left(a_{1}-1\right)}$.
From $00 \\ \left(a_{1}-1\right) n-a_{1}>a_{1}-1\end{array}, \therefore n>2+\frac{1}{a_{1}-1}, \because 1+\frac{1}{m}<a_{1}<\frac{m}{m-1}\right.$,
$\therefore m+1<2+\frac{1}{a_{1}-1}<m+2$, taking $N=m+2$, then when $n \geqslant N$, it always holds that $0<a_{n}<1$.