1. (20 points) Given that $x_{1}$ and $x_{2}$ are the two real roots of the quadratic equation in $x$
$$
x^{2}+(3 a-1) x+2 a^{2}-1=0
$$

such that
$$
\left(3 x_{1}-x_{2}\right)\left(x_{1}-3 x_{2}\right)=-80
$$

holds. Find all possible values of the real number $a$.
Three, 1. From the given conditions,
$$
\begin{array}{l}
\Delta=(3 a-1)^{2}-4\left(2 a^{2}-1\right) \\
=a^{2}-6 a+5 \geqslant 0 .
\end{array}
$$

Solving this, we get $a \geqslant 5$ or $a \leqslant 1$.
From the relationship between roots and coefficients, we know
$$
\begin{array}{l}
x_{1}+x_{2}=-(3 a-1), x_{1} x_{2}=2 a^{2}-1 . \\
\text { Then }\left(3 x_{1}-x_{2}\right)\left(x_{1}-3 x_{2}\right) \\
=3\left(x_{1}^{2}+x_{2}^{2}\right)-10 x_{1} x_{2} \\
=3\left(x_{1}+x_{2}\right)^{2}-16 x_{1} x_{2} \\
=3(3 a-1)^{2}-16\left(2 a^{2}-1\right) \\
=-5 a^{2}-18 a+19=-80 .
\end{array}
$$

Solving this, we get $a=3$ (discard) or $a=-\frac{33}{5}$.