Which are the numbers?

Observe that $8 = 2^3 = \left(4^{1/2}\right)^3 = 4^{2/3} = \sqrt[2]{4}^3$ Thus, $\boxed{n = 3,q=2}$ .

We need to test with the given values... If $q=2$ & $n=3$ , (\sqrt[2]{4})^3 = 2^3 =\boxed{\color\red{8}}

Consider that $(\sqrt[q]{4})^n=$ $4^{\frac{n}{q}}$

$4^{\frac{n}{q}}=$ $8$

$\frac{n}{q}=$ $Log_4(8)=$ $Log_2{(\sqrt{8})}$

$Log_2{(2^{\frac{3}{2}})}=$ $\frac{n}{q}$

So, $\frac{n}{q}=$ $\frac{3}{2}$ and $n=3, q=2$

The expression rules says: Radiciation first, potentiation second, and 2³ = 8, then: 3=n and 2=q.