$f\left(x\right)=\left\{\begin{array}{ll}{\left(1+\left|\sin \mathrm{x}\right|\right)}^{\frac{\mathrm{p}}{\left|\sin \text{ x}\right|}}& \frac{-\mathrm{\pi }}{6}<x<0\\ {\mathrm{e}}^{\frac{\sin 2\mathrm{x}}{\sin 3\mathrm{x}}}& 0<x<\frac{\mathrm{\pi }}{6}\end{array}\right.$ ,  $f\left(0\right)=q$ is continuous at $\mathrm{x}=0$ then 

$\mathrm{f}\left(\mathrm{x}\right) = \left\{\begin{array}{ll}\begin{array}{c}{\left(1+\left|\sin x\right|\right)}^{\frac{\mathrm{p}}{\left|\mathrm{sinx}\right|}}\end{array}& \mathrm{if}\frac{-\mathrm{\pi }}{6}<x<0\\ {\mathrm{e}}^{\frac{\sin 2\mathrm{x}}{\sin 3\mathrm{x}}}& \mathrm{ }\mathrm{if}\mathrm{ }0<\mathrm{x}<\frac{\mathrm{\pi }}{6}\\ \mathrm{q}& \mathrm{ }\mathrm{if}\mathrm{ }\mathrm{x}=0\end{array}\right.$is continuous at x = 0

$\therefore \mathrm{LHL}=\mathrm{RHL}=\mathrm{f}\left(\mathrm{o}\right)$

$\underset{\mathrm{x}\to 0^{-}}{\lim }\mathrm{ }{\left(1+\left|\mathrm{sinx}\right|\right)}^{\frac{\mathrm{p}}{\left|\mathrm{sinx}\right|}}=\underset{\mathrm{x}\to 0^{+}}{\lim }{\mathrm{e}}^{\frac{\sin 2\mathrm{x}}{\sin 3\mathrm{x}}} =\mathrm{q}$

$=\underset{\mathrm{x}\to 0^{+}}{\lim }\frac{\sin 2\mathrm{x}}{\mathrm{x}}\mathrm{ }\frac{x}{\sin 3\mathrm{x}}=\mathrm{q}$

${\mathrm{e}}^{\mathrm{p}}={\mathrm{e}}^{\frac{2}{3}}=\mathrm{q}\Rightarrow \mathrm{p}=\frac{2}{3},\mathrm{ }\mathrm{q}={\mathrm{e}}^{\frac{2}{3}}$

Option (1) is correct