13. Given that $a$ and $b$ are constants, and $a \neq 0$, the function $f(x)=\frac{x}{a x+3}$ satisfies $f(2)=1$, and the value of $x$ that makes $f(x)=x$ is unique. (1) Find the values of $a$ and $b$; (2) Let $x_{1}=1, x_{n}=$ $f\left(x_{n-1}\right)(n=2,3, \cdots)$, find the general term formula of the sequence $\left\{x_{n}\right\}$.
13. (1) From $f(2)=1$ we get $2a+b=2$, and the equation $f(x)=x$ has only one solution, i.e., $ax^2+(b-1)x=0$. It has only one solution, which must mean $b-1=0$, so $b=1, a=\frac{1}{2} \cdot f(x)=\frac{2x}{x+2}$;
(2) $x_{n}=\frac{2x_{n-1}}{x_{n-1}+2}$, thus $\frac{1}{x_{n}}=\frac{1}{x_{n-1}}+\frac{1}{2}$, hence $\left\{\frac{1}{x_{n}}\right\}$ is an arithmetic sequence with the first term $\frac{1}{x_{1}}=1$ and common difference $\frac{1}{2}$, so $\frac{1}{x_{n}}=1+(n-1) \frac{1}{2}=\frac{1}{2}(n+1)$. Therefore, $x_{n}=\frac{2}{n+1}\left(n \in \mathbf{N}_{+}\right)$.