7.  Find all positive real numbers $a$ such that the equation
$$
a^{2} x^{2}+a x+1-13 a^{2}=0
$$

has two integer roots. $\qquad$
$7.1, \frac{1}{3}, \frac{1}{4}$.
Let the two integer roots be $\alpha$ and $\beta$. Then, from $\alpha+\beta=-\frac{1}{a}$ being an integer, we know that the positive number $a=\frac{1}{n}$ (where $n$ is a positive integer).
Thus, the original equation can be written as
$$
x^{2}+n x+n^{2}-13=0 \text {. }
$$

By $\Delta=n^{2}-4\left(n^{2}-13\right)=52-3 n^{2} \geqslant 0$
$$
\Rightarrow n^{2} \leqslant \frac{52}{3} \text {. }
$$

Therefore, $n$ can take $1, 2, 3, 4$.
When $n=1$, the original equation is $x^{2}+x-12=0$, which has two integer roots;

When $n=2$, the original equation is $x^{2}+2 x-9=0$, which has no integer roots;

When $n=3$, the original equation is $x^{2}+3 x-4=0$, which has two integer roots;

When $n=4$, the original equation is $x^{2}+4 x+3=0$, which has two integer roots.
Thus, $a$ can take $1, \frac{1}{3}, \frac{1}{4}$.