Given,

$\tan 15°$ and $\tan 30°$ are the roots of the equation $x^2+px+q=0$,

So, sum of roots will be $\tan 15°+\tan 30°=-p ....\left(1\right)$

And product of roots will be $\tan 15°\tan 30°=q ....\left(2\right)$

Now we know that $\tan \left(A-B\right)=\frac{\tan A-\tan B}{1+\tan A\tan B}$

Using the above formula we get,

$\tan 15°=\frac{\tan 45°-\tan 30°}{1+\tan 45°\times \tan 30°}$

$\Rightarrow \tan 15°=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}=\frac{\sqrt{3}-1}{\sqrt{3}+1} ....\left(3\right)$

Now multiplying both the equation we get,

$-pq=\left(\tan 15°+\tan 30°\right)\left(\tan 15°\times \tan 30°\right)$

Now putting the value from equation $\left(3\right)$ we get,

$\Rightarrow -pq=\left(\frac{\sqrt{3}-1}{\sqrt{3}+1}+\frac{1}{\sqrt{3}}\right)\left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\times \frac{1}{\sqrt{3}}\right)$

$\Rightarrow -pq=\left(\frac{3-\sqrt{3}+\sqrt{3}+1}{\sqrt{3}\left(\sqrt{3}+1\right)}\right)\left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\times \frac{1}{\sqrt{3}}\right)$

$\Rightarrow -pq=\left(\frac{4\left(\sqrt{3}-1\right)}{3{\left(\sqrt{3}+1\right)}^2}\right)$

$\Rightarrow -pq=\left(\frac{4\sqrt{3}-4}{3\left(3+1+2\sqrt{3}\right)}\right)$

$\Rightarrow -pq=\left(\frac{2\sqrt{3}-2}{3\left(2+\sqrt{3}\right)}\right)$

$\Rightarrow -pq=\left(\frac{2\sqrt{3}-2}{3\left(2+\sqrt{3}\right)}\times \frac{\left(2-\sqrt{3}\right)}{\left(2-\sqrt{3}\right)}\right)$

$\Rightarrow -pq=\left(\frac{\left(2\sqrt{3}-2\right)\left(2-\sqrt{3}\right)}{3\times 1}\right)$

$\Rightarrow pq=\frac{10-6\sqrt{3}}{3}$