1. For the system of equations $\left\{\begin{array}{l}x^{2}+2 a y=5, \\ y-x=6 a\end{array}\right.$ to have positive integer solutions, the value of $a$ is . $\qquad$
2.1. $\frac{1}{2}$ or $\frac{1}{6}$.
From the second equation, we get $a=\frac{y-x}{6}$, substituting into the first equation and rearranging yields
$$
y^{2}-x y+3 x^{2}-15=0 \text {. }
$$

If this equation has positive integer solutions, then $\Delta=60-11 x^{2}$ must be a perfect square.
It is easy to see that $x=1,2$. Thus, $y=4,3$.
Therefore, we can find $a=\frac{1}{2}$ or $\frac{1}{6}$.