1. Let us set $S = |\sqrt{1 + \sin 2x} - \sqrt{1 - \sin 2x}|$. Observe that $S^2 = 2 - 2 \sqrt{1 - \sin^2 2x} = 2 - 2|\cos 2x| \leq 2$, implying $S \leq \sqrt{2}$. Thus the righthand inequality holds for all $x$. It remains to investigate the left-hand inequality. If $\pi / 2 \leq x \leq 3 \pi / 2$, then $\cos x \leq 0$ and the inequality trivially holds. Assume now that $\cos x > 0$. Then the inequality is equivalent to $2 + 2 \cos 2x = 4 \cos^2 x \leq S^2 = 2 - 2|\cos 2x|$, which is equivalent to $\cos 2x \leq 0$, i.e., to $x \in [\pi / 4, \pi / 2] \cup [3 \pi / 2, 7 \pi / 4]$. Hence the solution set is $\pi / 4 \leq x \leq 7 \pi / 4$.