a) By the mean value theorem there is $c\in(-\pi,\pi)$ such that $$ F'(c)={F(\pi )-F(-\pi)\over \pi -(-\pi)}={{\pi ^3\over 2}-{-\pi ^3\over 2}\over \pi -(-\pi)}={\pi ^3\over 2\pi}={\pi ^2\over 2}$$
b) For $x\in(0,\pi/3)$ we have $$2=2+\tan 0\le 2+\tan x\le 2+\tan(\pi/3)=2+\sqrt{3} \\\; \\ \implies {x^3\over2+\sqrt{3}}\le {x^3\over2+\tan x}\le {x^3\over2} \\ \;\\ \implies \int _0^{\pi/3}{x^3\over2+\sqrt{3}}dx\le \int _0^{\pi/3}{x^3\over2+\tan x}dx\le \int _0^{\pi/3}{x^3\over2}dx \\ \; \\ \implies {\pi\over4}{\pi^3\over81(2+\sqrt{3})}={\pi^4\over81\times 4(2+\sqrt{3})} ={x^4\over4(2+\sqrt{3})}\Big| _0^{\pi/3} \le \int _0^{\pi/3}{x^3\over2+\tan x}dx \\ \; \\ \le {x^4\over8}\Big| _0^{\pi/3}={\pi^4\over 81\times 8}={\pi^3\over 162}{\pi\over 4}<{\pi^3\over 162}$$
Since $\pi/4<1$ I suspect the lower bound is a typo. Or the power of $x$ is a typo, because we have
$$2=2+\tan 0\le 2+\tan x\le 2+\tan(\pi/3)=2+\sqrt{3} \\\; \\ \implies {x^2\over2+\sqrt{3}}\le {x^2\over2+\tan x}\le {x^2\over2} \\ \;\\ \implies \int _0^{\pi/3}{x^2\over2+\sqrt{3}}dx\le \int _0^{\pi/3}{x^2\over2+\tan x}dx\le \int _0^{\pi/3}{x^2\over2}dx \\ \; \\ \implies {\pi^3\over81(2+\sqrt{3})} ={x^3\over3(2+\sqrt{3})}\Big| _0^{\pi/3} \le \int _0^{\pi/3}{x^2\over2+\tan x}dx \le {x^3\over6}\Big| _0^{\pi/3}={\pi^3\over 162}$$
Martin
