Let x=e^(2∏/13i), a primitive 13th root of unity.
Find a subfield K of Q(X) with (Q(x):K)=3.
Find a subfield L of Q(X) with (Q(x):L)=4.
Let $\displaystyle \zeta$ be the primitive 13th roots of unity. We see that $\displaystyle \text{Gal}(\mathbb{Q}(\zeta), \mathbb{Q})=(\mathbb{Z}/13\mathbb{Z})^\times \cong \mathbb{Z}/12\mathbb{Z}$, where the generator of this cyclic group is $\displaystyle \sigma:\zeta \mapsto \zeta^2$.
You need to find the subgroup of the order 3 and order 4 for the above group and correspond them to the subfields of $\displaystyle \mathbb{Q}(\zeta)$.
The subgroup of order 3 for $\displaystyle \text{Gal}(\mathbb{Q}(\zeta), \mathbb{Q})$ is generated by $\displaystyle \sigma^4$.
Thus $\displaystyle K=\mathbb{Q}(\zeta + \sigma^4(\zeta) + \sigma^8(\zeta))$ $\displaystyle = \mathbb{Q}(\zeta + \zeta^{2^4} + \zeta^{2^8}) = \mathbb{Q}(\zeta + \zeta^3 + \zeta^9)$, where $\displaystyle [\mathbb{Q}(\zeta):K ] = 3$.
You can find the L exactly in the same way.