$1)y=\sqrt{2+\tan\left(x+\dfrac{1}{x}\right)}\\ y'=\dfrac{1}{2\sqrt{2+\tan\left(x+\dfrac{1}{x}\right)}}.\left(2+\tan\left(x+\dfrac{1}{x}\right)\right)'\\ =\dfrac{1}{2\sqrt{2+\tan\left(x+\dfrac{1}{x}\right)}}.\left[\tan^2\left(x+\dfrac{1}{x}\right)+1\right].\left(x+\dfrac{1}{x}\right)'\\ =\dfrac{\left[\tan^2\left(x+\dfrac{1}{x}\right)+1\right].\left(1-\dfrac{1}{x^2}\right)}{2\sqrt{2+\tan\left(x+\dfrac{1}{x}\right)}}\\ 2)y=\cos\left(\dfrac{2\pi}{3}+2x\right)\\ y'=-2\sin\left(\dfrac{2\pi}{3}+2x\right)\\ y'=0\\ \Leftrightarrow -2\sin\left(\dfrac{2\pi}{3}+2x\right)=0\\ \Leftrightarrow \sin\left(\dfrac{2\pi}{3}+2x\right)=0\\ \Leftrightarrow \dfrac{2\pi}{3}+2x=k\pi(k \in \mathbb{Z})\\ \Leftrightarrow 2x=k\pi-\dfrac{2\pi}{3}(k \in \mathbb{Z})\\ \Leftrightarrow x=k\dfrac{\pi}{2}-\dfrac{\pi}{3}(k \in \mathbb{Z})$
