13. Given the inequality $|a x-3| \leqslant b$ has the solution set $\left[-\frac{1}{2}, \frac{7}{2}\right]$. Then $a+b=$ $\qquad$ .
13.6.

From $|a x-3| \leqslant b$, we get $3-b \leqslant a x \leqslant 3+b$. It is easy to see that $a \neq 0$, so $\frac{3-b}{a}+\frac{3+b}{a}=-\frac{1}{2}+\frac{7}{2}$. Solving this, we get $a=2$.
Therefore, $\frac{3-b}{a}=-\frac{1}{2}$. Solving this, we get $b=4$. Hence, $a+b=6$.