$\cos \theta =\frac{-\sqrt{3}}{2}$and $\sin \alpha =\frac{-3}{5}$ where $\theta \mathrm{and} \alpha$ does not lie on the third quadrant,  then $5\cos \alpha +\sqrt{3}\tan \theta =$

$\cos \theta =\frac{-\sqrt{3}}{2}$

$\cos \theta \mathrm{is} \mathrm{negative} \mathrm{in} {\mathrm{II}}^{\mathrm{nd}} , {\mathrm{III}}^{\mathrm{rd}} \mathrm{quadrants}$

$\theta$ does not belongs to third quadrant

$\therefore \theta$ belong to $2^{\mathrm{nd}}$ quadrant

$\tan \theta =\frac{-1}{\sqrt{3}}$

$\sin \alpha =\frac{-3}{5}\Rightarrow \alpha$does not belongs to $3^{\mathrm{rd}}$ quadrant

$\Rightarrow \alpha$ belong to $4^{\mathrm{th}}$ quadrant

$\cos \alpha =\frac{4}{5}$

$5\cos \alpha +\sqrt{3}\tan \theta =5\left(\frac{4}{5}\right)+\sqrt{3}\left(-\frac{1}{\sqrt{3}}\right)$

$=4-1=3$