$I={\int }_{\frac{\pi }{4}}^{\frac{\pi }{3}}\left(\frac{8\sin x-\sin 2x}{x}\right)dx$. Then

Let, $f\left(x\right)=8\sin x-\sin 2x$

$\Rightarrow f^{\text{'}}\left(x\right)=8\cos x-2\cos 2x$

$\Rightarrow f^{\text{'}\text{'}}\left(x\right)=-8\sin x+4\sin 2x$

$\Rightarrow f^{\text{'}\text{'}}\left(x\right)=-8\sin x\left(1-\cos x\right)$

So, $f^{\text{'}}\left(x\right)<0 \text{when }x\in \left(\frac{\pi }{4},\frac{\pi }{3}\right)$

Hence $f^{\text{'}}\left(x\right)$ is decreasing function in given domain,

Now, $f^{\text{'}}\left(\frac{\pi }{3}\right)<f^{\text{'}}\left(x\right)<f^{\text{'}}\left(\frac{\pi }{4}\right)$

$\Rightarrow 5<f^{\text{'}}\left(x\right)<\frac{8}{\sqrt{2}}$

$\Rightarrow 5<f^{\text{'}}\left(x\right)<4\sqrt{2}$

Now integrating all with respect to $dx$

$\int 5dx<\int f^{\text{'}}\left(x\right)dx<\int 4\sqrt{2}dx$

$\Rightarrow 5x<f\left(x\right)<4\sqrt{2}x$

$\Rightarrow 5<\frac{f\left(x\right)}{x}<4\sqrt{2}$

Again integrating with ${\int }_{\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{3}}$

We get, ${\int }_{\frac{\pi }{4}}^{\frac{\pi }{4}}5<\int \frac{f\left(x\right)}{x}<{\int }_{\frac{\pi }{4}}^{\frac{\pi }{3}}4\sqrt{2}$

$\Rightarrow {\int }_{\frac{\pi }{4}}^{\frac{\pi }{3}}5<\int \frac{8\sin x-\sin 2x}{x}<{\int }_{\frac{\pi }{4}}^{\frac{\pi }{3}}4\sqrt{2}$

$\Rightarrow \frac{5\pi }{12}<I<\frac{\sqrt{2}\pi }{3}$