13. Given the function $f(x)=\left|1-\frac{1}{x}\right|$.
(1) Does there exist real numbers $a, b (a<b)$ such that the domain and range of the function $f(x)$ are both $[a, b]$? If they exist, find the values of $a, b$; if not, explain the reason.
(2) If there exist real numbers $a, b (a<b)$ such that the domain of the function $f(x)$ is $[a, b]$ and the range is $[m a, m b]$ $(m \neq 0)$, find the range of the real number $m$.
Three, 13. (1) There do not exist real numbers $a, b (a < b)$ such that $f(x) = mx$ for all $x \in (a, b)$, where $m > 0, a > 0$.

Given:
$$
f(x)=\left\{\begin{array}{l}
1-\frac{1}{x}, x \geqslant 1 ; \\
\frac{1}{x}-1, 0 < x < 1.
\end{array}\right.
$$

By analogy with the solution to (1), we know that when $a, b \in (0, 1)$ or $a \in (0, 1), b \in [1, +\infty)$, there do not exist such $a, b$.

Only when $a, b \in [1, +\infty)$, $f(x) = 1 - \frac{1}{x}$ is an increasing function on $[1, +\infty)$. Therefore,
$$
\left\{\begin{array}{l}
f(a) = ma, \\
f(b) = mb,
\end{array}\right.
$$
which means
$$
\left\{\begin{array}{l}
1 - \frac{1}{a} = ma, \\
1 - \frac{1}{b} = mb.
\end{array}\right.
$$

Thus, $a, b$ are the two real roots greater than 1 of the equation $mx^2 - x + 1 = 0$. Therefore,
$$
\left\{\begin{array}{l}
\Delta = 1 - 4m > 0, \\
x = \frac{1 \pm \sqrt{1 - 4m}}{2m} > 1,
\end{array}\right.
$$
which only requires
$$
\left\{\begin{array}{l}
m > 0, \\
1 - 4m > 0, \\
1 - \sqrt{1 - 4m} > 2m.
\end{array}\right.
$$

Solving this, we get $0 < m < \frac{1}{4}$.
Therefore, the range of $m$ is $0 < m < \frac{1}{4}$.