2. Given the function $y=x^{2}+6 a x-a$ intersects the $x$-axis at two distinct points $\left(x_{1}, 0\right)$ and $\left(x_{2}, 0\right)$, and
$$
\begin{array}{l}
\frac{a}{\left(1+x_{1}\right)\left(1+x_{2}\right)}-\frac{3}{\left(1-6 a-x_{1}\right)\left(1-6 a-x_{2}\right)} \\
=8 a-3 .
\end{array}
$$

Then the value of $a$ is $\qquad$
2. $\frac{1}{2}$.

From $\Delta=36 a^{2}+4 a>0$, we get $a>0$ or $a<-\frac{1}{9}$. According to the problem,
$$
\begin{array}{l}
y=x^{2}+6 a x-a=\left(x-x_{1}\right)\left(x-x_{2}\right) . \\
\text { Then }\left(1+x_{1}\right)\left(1+x_{2}\right)=f(-1)=1-7 a, \\
\left(1-6 a-x_{1}\right)\left(1-6 a-x_{2}\right) \\
=f(1-6 a)=1-7 a .
\end{array}
$$

Thus, $\frac{a-3}{1-7 a}=8 a-3$.
Solving this, we get $a=\frac{1}{2}$ or $a=0$ (discard).