(13) Given $a>0$, the solution set of the inequality $\left|\frac{2 x-3-2 a}{x-a}\right| \leqslant 1$ is
(13) The original inequality is equivalent to
$$
\left\{\begin{array}{l}
\frac{2 x-3-2 a}{x-a} \leqslant 1, \\
\frac{2 x-3-2 a}{x-a} \geqslant-1,
\end{array}\right.
$$

Rearranging and finding a common denominator, we get
$$
\left\{\begin{array}{l}
\frac{x-(a+3)}{x-a} \leqslant 0, \\
\frac{3[x-(a+1)]}{x-a} \geqslant 0 .
\end{array}\right.
$$

Given that $a>0$, solving (1) yields $a<x \leqslant a+3$, and solving (2) yields $x \geqslant a+1$ or $x<a$.
Therefore, the solution set of the original inequality is $\{x \mid a+1 \leqslant x \leqslant a+3\}$.