1. (16 points) Suppose the equation $\left|x^{2}+a x\right|=4$ has only 3 distinct real roots. Find the value of $a$ and the corresponding 3 roots.
Three, 1. The equation $\left|x^{2}+a x\right|=4$ is equivalent to the following two equations:
$x^{2}+a x-4=0$,
$x^{2}+a x+4=0$.
Let $x_{0}$ be a root of equation (1), then $x_{0}^{2}+a x_{0}-4=0$.
Also, $x_{0}^{2}+a x_{0}+4=x_{0}^{2}+a x_{0}-4+8=8 \neq 0$,
Therefore, $x_{0}$ is definitely not a root of equation (2), so equations (1) and (2) have no common roots.

Since there are only 3 distinct real roots, one of the equations (1) or (2) must have a repeated root.
$\Delta_{1}=a^{2}+16 \geqslant 0, \Delta_{2}=a^{2}-16 \geqslant 0$.
Since $\Delta_{1}>0$, it can only be that $\Delta_{2}=0$, i.e., $a= \pm 4$.
When $a=4$, the roots of equation (2) are -2, and the roots of equation (1) are $-2 \pm 2 \sqrt{2}$.

When $a=-4$, the roots of equation (2) are 2, and the roots of equation (1) are $2 \pm 2 \sqrt{2}$.
In summary, when $a=4$, the 3 roots are $-2,-2 \pm 2 \sqrt{2}$;
When $a=-4$, the 3 roots are $2, 2 \pm 2 \sqrt{2}$.