3.I.1. The domain of the inequality is the union of the intervals $(-\infty; -2]$ and $(4; +\infty)$.

Suppose that $x>4$. In this case, the equation transforms into $(x+2)(x-4)+4 \sqrt{(x+2)(x-4)}=12$. By making the substitution $t=$ $\sqrt{(x+2)(x-4)}$, we obtain the equation $t^{2}+4 t-12=0$, from which $t=-6$ or $t=2$. Since $t \geqslant 0$, then $\sqrt{(x+2)(x-4)}=2$, or $x^{2}-2 x-12=0$, from which $x=1 \pm \sqrt{13}$. Since $x>4$, then $x=1+\sqrt{13}$.

Now let $x \leqslant-2$. In this case, $(x-4) \sqrt{\frac{x+2}{x-4}}=-\sqrt{(x+2)(x-4)}$. The same substitution as above reduces the equation to $t^{2}-4 t-12=0$, so $\sqrt{(x+2)(x-4)}=6$, from which $x^{2}-2 x-44=0$, thus $x=1 \pm 3 \sqrt{5}$. Since $x \leqslant-2$, then $x=1-3 \sqrt{5}$.

Answer: $\{1+\sqrt{13}, 1-3 \sqrt{5}\}$.
1. Solve the equation $(x+2)(x-4)+4(x-4) \sqrt{\frac{x+2}{x-4}}=12$.