Answer
a. $y(\frac{\pi}{3})=2\sqrt 3$, absolute minimum.
b. See graph; absolute minimum at $(\frac{\pi}{3},2\sqrt 3)$
Work Step by Step
a. Given $y=tan(x)+3cot(x), 0\lt x\lt\pi/2$, we have $y'=sec^2(x)-3 csc^2(x)$. Letting $y'=0$, we have $tan(x)=\pm\sqrt 3$ which gives $x=\frac{\pi}{3}$ in $0\lt x\lt\pi/2$. We have $y(\frac{\pi}{3})=tan(\frac{\pi}{3})+3cot(\frac{\pi}{3})=2\sqrt 3$. Checking signs of $y'$ across $x=\frac{\pi}{3}$, we have $(0)..(-)..(\frac{\pi}{3})..(+)..(\pi)$, indicating $y(\frac{\pi}{3})=2\sqrt 3$ is an absolute minimum.
b. See graph. We can identify an absolute minimum at $(\frac{\pi}{3},2\sqrt 3)$, which agrees with the result from part (a).
