Prove that $\frac{\sqrt{3}}{8}<\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{\sin x}{x}\,dx<\frac{\sqrt{2}}{6}$ 
Prove that $$\frac{\sqrt{3}}{8}<\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{\sin x}{x}\,dx<\frac{\sqrt{2}}{6}$$

My try: Using $$\displaystyle \sin x<x$$ and $$\frac{\sin x-0}{x-0}>\frac{1-0}{\frac{\pi}{2}-0}=\frac{2}{\pi}$$
So we get $$\frac{2}{\pi}<\frac{\sin x}{x}<1$$
So we get  $$\frac{2}{\pi}\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}1\,dx<\frac{\sin x}{x}<1\cdot \int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\,$$
But this is not what I have to prove here.

This is not an answer but it is too long for a comment.
I cannot resist the pleasure of reporting again the magnificent approximation proposed by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician.$$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$ (see here).
So, as an approximation $$\int\frac{\sin(x)}x dx\approx-2 \left(\log \left(4 x^2-4 \pi  x+5 \pi ^2\right)+\tan
   ^{-1}\left(\frac{1}{2}-\frac{x}{\pi }\right)\right)$$ which gives $$\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{\sin x}{x}dx\approx 2 \log \left(\frac{153}{148}\right)+\tan ^{-1}\left(\frac{100}{621}\right)\approx 0.226111$$ while the exact solution is $$\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{\sin x}{x}dx=\text{Si}\left(\frac{\pi }{3}\right)-\text{Si}\left(\frac{\pi }{4}\right)\approx 0.226483$$ your bounds being $\approx 0.216506$ and $\approx 0.235702$
