a,
$T=1-\cos^2x+\cos^4x$
Đặt $t=\cos^2x\Rightarrow T=t^2-t+1$
$T=t^2-2t.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=(t-\dfrac{1}{2})^2+\dfrac{3}{4}\ge \dfrac{3}{4}$
$\Rightarrow \min T=\dfrac{3}{4}\Leftrightarrow t=\cos^2x=\dfrac{1}{2}$
$\Leftrightarrow \dfrac{1+\cos 2x}{2}=\dfrac{1}{2}$
$\Leftrightarrow \cos 2x=0$
$\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}$
b,
$y=\sin x-\cos x$
$=\sqrt2 \sin(x-\dfrac{\pi}{4})$
$-1\le \sin(x-\dfrac{\pi}{4})\le 1$
$\Leftrightarrow -\sqrt2 \le y \le \sqrt2$
$\Rightarrow \min y=-\sqrt2$
