13. Let $a>0$, solve the inequality $\sqrt{2 a x-a^{2}}>1-x$ for $x$.
13. The original inequality is equivalent to (1) $\left\{\begin{array}{l}1-x \geqslant 0 \\ 2 a x-a^{2}>(1-x)^{2}\end{array}\right.$ or (2) $\left\{\begin{array}{l}1-x<1 \\ x \geqslant \frac{a}{2},\end{array}\right.$ when $\frac{a}{2} \geqslant 1$ i.e., $a \geqslant 2$, $x>\frac{a}{2}$; when $0<a<2$, $1-x<1$ and $x \geqslant \frac{a}{2}$, thus $1+a-\sqrt{2 a}<x<1$. In summary: when $a \geqslant 2$, the solution set of the original inequality is $\left\{x \left\lvert\, x>\frac{a}{2}\right.\right\}$; when $0<a<2$, the solution set is $\{x \mid 1+a-\sqrt{2 a}<x<1\}$.