The given limit is

$l=\underset{n\to \infty }{\lim }\frac{1}{n}\left[\sin \frac{\pi }{4}+\sin \frac{\pi }{12}\left(3+\frac{1}{n}\right)+\sin \frac{\pi }{12}\left(3+\frac{2}{n}\right)+...+\sin \frac{\pi }{3}\right]$

$\Rightarrow l=\underset{n\to \infty }{\lim }\frac{1}{n}\left[\sin \frac{\pi }{4}+\sin \left(\frac{\mathrm{\pi }}{4}+\frac{\mathrm{\pi }}{12n}\right)+\sin \left(\frac{\mathrm{\pi }}{4}+\frac{2}{12n}\right)+...+\sin \left(\frac{\pi }{4}+\left(n\right)\frac{\mathrm{\pi }}{12n}\right)\right]$

We know that

$\sin \alpha +\sin \left(\alpha +\beta \right)+\sin \left(\alpha +2\beta \right)+...+\sin \left(\alpha +\left(N-1\right)\beta \right)=\frac{\sin \left({\displaystyle \frac{N\beta }{2}}\right)\sin \left(\alpha +{\displaystyle \frac{\left(N-1\right)\beta }{2}}\right)}{\sin \left({\displaystyle \frac{\beta }{2}}\right)}.$

Here, $\alpha =\frac{\mathrm{\pi }}{4}, \beta =\frac{\mathrm{\pi }}{12n}, N=n+1.$

$\Rightarrow l=\underset{n\to \infty }{\lim }\frac{1}{n}\left[\frac{\sin \left({\displaystyle \frac{\left(n+1\right)\mathrm{\pi }}{24n}}\right)\sin \left({\displaystyle \frac{\mathrm{\pi }}{4}}+n\times {\displaystyle \frac{\mathrm{\pi }}{24n}}\right)}{\sin \left({\displaystyle \frac{\mathrm{\pi }}{24n}}\right)}\right]$

$\Rightarrow l=\underset{n\to \infty }{\lim }\frac{1}{n}\left[\frac{\sin \left({\displaystyle \frac{\mathrm{\pi }}{24}+\frac{\mathrm{\pi }}{24n}}\right)\sin \left({\displaystyle \frac{\mathrm{\pi }}{4}+\frac{\mathrm{\pi }}{24}}\right)}{\left({\displaystyle \frac{\mathrm{\pi }}{24n}}\right)\left({\displaystyle \frac{\sin \left({\displaystyle \frac{\mathrm{\pi }}{24\mathrm{n}}}\right)}{\left({\displaystyle \frac{\mathrm{\pi }}{24\mathrm{n}}}\right)}}\right)}\right]$

$\Rightarrow l=\underset{n\to \infty }{\lim }\left[\frac{24\sin \left({\displaystyle \frac{\mathrm{\pi }}{24}+\frac{\mathrm{\pi }}{24n}}\right)\sin \left({\displaystyle \frac{7\mathrm{\pi }}{24}}\right)}{\left({\displaystyle \mathrm{\pi }}\right)\left({\displaystyle \frac{\sin \left({\displaystyle \frac{\mathrm{\pi }}{24\mathrm{n}}}\right)}{\left({\displaystyle \frac{\mathrm{\pi }}{24\mathrm{n}}}\right)}}\right)}\right]$

Using $\underset{x\to 0}{\lim }\left(\frac{\sin x}{x}\right)=1$ and $\underset{n\to \infty }{\lim }\left(\frac{1}{n}\right)=0,$

$\Rightarrow l=\left[\frac{24\sin \left({\displaystyle \frac{\mathrm{\pi }}{24}}\right)\sin \left({\displaystyle \frac{7\mathrm{\pi }}{24}}\right)}{\mathrm{\pi }}\right]$

Using $2\sin A\sin B=\cos \left(A-B\right)\cos \left(A+B\right),$

$\Rightarrow l=\frac{12}{\mathrm{\pi }}\left[\cos \left(\frac{7\mathrm{\pi }}{24}-\frac{\mathrm{\pi }}{24}\right)-\cos \left(\frac{7\mathrm{\pi }}{24}+\frac{\mathrm{\pi }}{24}\right)\right]$

$\Rightarrow l=\frac{12}{\mathrm{\pi }}\left[\cos \left(\frac{\mathrm{\pi }}{4}\right)-\cos \left(\frac{\mathrm{\pi }}{3}\right)\right]$

$\Rightarrow l=\frac{12}{\mathrm{\pi }}\left[\frac{1}{\sqrt{2}}-\frac{1}{2}\right]$

$\Rightarrow l=\frac{12}{\mathrm{\pi }}\left[\frac{\sqrt{2}-1}{2}\right]=\frac{6\left(\sqrt{2}-1\right)}{\mathrm{\pi }}.$