Given, $\cos \theta =-\frac{\sqrt{3}}{2}<0$ and $\theta$ does not lie in third quadrant.

$\therefore \theta$ must be lying in $2^{nd}$ quadrant.

$\Rightarrow \tan \theta =-\frac{1}{\sqrt{3}}$ and $\cot \theta =-\sqrt{3} ...\left(\mathrm{i}\right)$

As, $\sin \alpha =-\frac{3}{5},$ thus $\alpha$ lies in $3^{rd}$ or $4^{th}$ quadrant.

If $\alpha$ is in third quadrant, then $\tan \alpha =\frac{3}{4}$ and $\cos \alpha =-\frac{4}{5}.$

$\therefore \frac{2\tan \alpha +\sqrt{3}\tan \theta }{{\cot }^2\theta +\cos \alpha }=\frac{2·\frac{3}{4}-\sqrt{3}·\frac{1}{\sqrt{3}}}{3-\frac{4}{5}}$

$=\frac{\frac{3}{2}-1}{3-\frac{4}{5}}=\frac{5}{22}.$

If $\alpha$ is in fourth quadrant, then $\tan \alpha =-\frac{3}{4}$ and $\cos \alpha =\frac{4}{5}.$

$\therefore \frac{2\tan \alpha +\sqrt{3}\tan \theta }{{\cot }^2\theta +\cos \alpha }=\frac{2·{\displaystyle \left(-\frac{3}{4}\right)}+\sqrt{3}·\left(-{\displaystyle \frac{1}{\sqrt{3}}}\right)}{3+{\displaystyle \frac{4}{5}}}$

$=\frac{\left(-\frac{3}{2}-1\right)}{\left(3+\frac{4}{5}\right)}=-\frac{5}{38}.$