Absolute-value inequalities

$-11<2x-3<11$, hence $-8<2x<14$ and $-4<x<7$.

$x^2-8\geq4$ or $x^2-8\leq-4$. Thus $x^2\geq12$ or $x^2\leq4$. Equivalently $|x|\geq\sqrt{12}$ or $|x|\leq\sqrt4$. Therefore $x\leq-\sqrt{12}$, $-2\leq x\leq2$, or $x\geq\sqrt{12}$.