13. (25 points) Given positive integers $x, y$ satisfy
$$
\frac{2}{x}+\frac{1}{y}=a\left(a \in \mathbf{N}_{+} \text {, and } x<y\right) \text {. }
$$
13. Given $x \geqslant 1, y \geqslant 2$, we have $\frac{2}{x}+\frac{1}{y} \leqslant 2+\frac{1}{2}$, i.e., $a \leqslant 2 \frac{1}{2}$.
Therefore, $a=1$ or 2.
When $a=1$, $\frac{2}{x}+\frac{1}{y}=1$.
If $x=1$, then $\frac{1}{y}=-1$, which contradicts that $y$ is a positive integer. Hence, $x \neq 1$.

If $x=2$, then $\frac{1}{y}=0$, which contradicts that $y$ is a positive integer. Hence, $x \neq 2$.

If $x \geqslant 3$, then $\frac{2}{x}+\frac{1}{y} \leqslant \frac{2}{3}+\frac{1}{4}<1$, which contradicts $\frac{2}{x}+\frac{1}{y}=1$. Hence, $x<3$.
In summary, $a \neq 1$.
When $a=2$, $\frac{2}{x}+\frac{1}{y}=2$.
If $x=1$, then $\frac{1}{y}=0$, which contradicts that $y$ is a positive integer. Hence, $x \neq 1$.

If $x \geqslant 3$, then $\frac{2}{x}+\frac{1}{y} \leqslant \frac{2}{3}+\frac{1}{4}<1$, which contradicts $\frac{2}{x}+\frac{1}{y}=2$. Hence, $x<3$.
Therefore, $1<x<3$. Thus, $x=2, y=1$.