Let $f\left(x\right)=\left\{\begin{array}{cc}{\left(1+\left|\sin x\right|\right)}^{\frac{l}{\left|\sin x\right|}},& -\frac{\pi }{6}<x<0\\ e^{\frac{\tan 2x}{\tan 3x}},& 0<x<\frac{\pi }{6}\\ m,& x=0\end{array}\right.$ is continuous at $x=0$. Then, the values of $l$ and $m$ are

We have,
$\underset{h\to 0^{+}}{\lim }f\left(0-h\right)=\underset{h\to 0^{+}}{\lim }{\left[1+\left|\sin \left(-h\right)\right|\right]}^{\frac{l}{\left|\sin -h\right|}}$
$=\underset{h\to 0^{+}}{\lim }{\left(1+\sin h\right)}^{\frac{l}{\sin h}}$
$={\underset{h\to 0^{+}}{\lim }\left[{\left(1+\sin h\right)}^{\frac{1}{\sin h}}\right]}^l=e^l$
$\underset{h\to 0^{+}}{\lim }f\left(0+h\right)=\underset{h\to 0^{+}}{\lim }{e}^{\frac{\tan 2h}{\tan 3h}}$
$=e^{\underset{h\to 0^{+}}{\lim }\left(\frac{\tan 2h}{2h}\times \frac{2}{3}\times \frac{3h}{\tan 3h}\right)}=e^{1\times \frac{2}{3}\times 1}=e^{\frac{2}{3}}$

Also, $f\left(0\right)=m$
For $f\left(x\right)$ to be continuous at $x=0,$ we must have
$$\begin{gathered} \underset{h\to 0^{+}}{\lim }f\left(0-h\right)=\underset{h\to 0^{+}}{\lim }f\left(0+h\right)=f\left(0\right) \\ \Rightarrow e^l=e^{\frac{2}{3}}=m \end{gathered}$$
$\Rightarrow l=\frac{2}{3}$ and $m=e^{\frac{2}{3}}$