Since, function is continuous at $x=0$

$f\left(0^{+}\right)=f\left(0^{-}\right)=f\left(0\right)$

Now, $f\left(0^{+}\right)=\underset{x\to 0^{+}}{\lim }q\left(\frac{\sqrt{x^2+x}-\sqrt{x}}{x^{{\displaystyle \frac{3}{2}}}}\right)$

$\Rightarrow f\left(0^{+}\right)=q\left(\underset{x\to 0^{+}}{\lim }\frac{\left(x^2+x-x\right)}{x^{{\displaystyle \frac{3}{2}}}}.\left(\frac{1}{\sqrt{x^2+x}+\sqrt{x}}\right)\right)$

$\Rightarrow f\left(0^{+}\right)=q\left(\underset{x\to 0^{+}}{\lim }\frac{x^2}{x\sqrt{x}}.\frac{1}{\sqrt{x}\left(\sqrt{x+1}+1\right)}\right)=\frac{q}{2}$

$f\left(0^{-}\right)=\underset{x\to 0^{-}}{\lim }\frac{\tan \left(2p-7\right)x+\tan 3x}{x}$

$\Rightarrow f\left(0^{-}\right)=\left(2p-7\right)+3=2p-4$

Since function is continuous at $x=0$, $f\left(0^{+}\right)=f\left(0\right)$

$\Rightarrow \frac{q}{2}=p-q$

$\Rightarrow 3q=2p$

$\Rightarrow \frac{q}{p}=\frac{2}{3}$