So $E=GF(3^{12})$ is an extension of $F=GF(3^4)$, and I wish to know how many intermediate fields $F \subsetneq K \subsetneq E$ there are.
By a result in Escofier's Galois Theory I have that $G={\rm Gal}(E/F)$ is cyclic and of order $3$, so $G\cong \mathbb{Z}_3$. Since the fundamental theorem gives a bijective correspondence between intermediate fields $F \subset K \subset E$ and subgroups of $G$, and since $\mathbb{Z}_3$ has no proper non-trivial subgroups, there shouldn't be any intermediate fields $F \subsetneq K \subsetneq E$. Am I missing something or is this correct? It seemed a bit too easy, especially since I often get confused when doing Galois theory and expect to get stuck.
Yes, your argument is correct. In general, the lattice of subfields of $GF(p^n)$ is isomorphic to the lattice of divisors of $n$. For example, the lattice of subfields of $GF(3^{12})$ will look exactly like the lattice of divisors of 12 which contains the elements 12, 6, 4, 3, 2 and 1. In this lattice, there will not be any elements between 4 and 12 because the ratio 12/4 is a prime.