e,
$-1\le \sin x^2\le 1$
$\to -1\le -\sin x^2\le 1$
$\to 0\le 1-\sin x^2\le 2$
$\to 0\le \sqrt{1-\sin x^2}\le \sqrt2$
$\to -1\le y\le \sqrt2-1$
Vậy: $\min y=-1; \max y=\sqrt2-1$
f,
$-1\le \sin\sqrt{x}\le 1$
$\to -4\le y\le 4$
Vậy $\min y=-4; \max y=4$