$y=\sqrt{5-2\sin^2x.\cos^2x}$
$=\sqrt{5-\dfrac{1}{2}\sin^22x}$
$=\sqrt{5-\dfrac{1-\cos4x}{4}}$
$=\sqrt{\dfrac{19}{4}+\dfrac{1}{4}\cos4x}$
$-1\le \cos4x\le 1$
$\Leftrightarrow \dfrac{3}{\sqrt2}\le y\le \sqrt5$
$\Rightarrow \min=\dfrac{3}{\sqrt2};\max=\sqrt5$