Hierarchy of complexity classes $\bigcup_{c > 0} \mathrm{Time}(2^{c \log^k n})$, w.r.t. $k$
Hint: Use the time hierarchy theorem.

The first step, naturally, is understanding the question, that is, the definitions involved. The class $\mathsf{QP}_k$ consists of all problems solvable in time $2^{O(\log^k n)}$. We could similarly define $\mathsf{P}_k$ to consist of all problems solvable in time $O(n^k)$, and $\mathsf{P}$ to be a union over all $\mathsf{P}_k$:
$$ \mathsf{P}_k = \bigcup_{c&gt;0} \mathsf{Time}(cn^k), \quad \mathsf{P} = \bigcup_{k&gt;0} \mathsf{P}_k = \bigcup_{c,k&gt;0} \mathsf{Time}(cn^k). $$
The class $\mathsf{QP} = \bigcup_{k&gt;0} \mathsf{QP}_k$ is known as quasi-polynomial time. The question asks you to show that the corresponding hierarchy $\mathsf{QP}_2 \subseteq \mathsf{QP}_3 \subseteq \cdots$ is strict, that is $\mathsf{QP}_2 \subsetneq \mathsf{QP}_3 \subsetneq \cdots$. The same happens with the hierarchy corresponding to $\mathsf{P}$, again using the time hierarchy theorem: $\mathsf{P}_1 \subsetneq \mathsf{P}_2 \subsetneq \cdots$.

Why can't the same approach be used to prove $\mathsf{P} \subsetneq \mathsf{NP}$? Because here we're comparing deterministic computation to non-deterministic computation with the same time bounds, and there is no counterpart of the time hierarchy theorem which helps us here. Moreover, the time hierarchy theorems, in both the deterministic and non-deterministic case, are proved using diagonalization, a technique which is known to be unable to separate $\mathsf{P}$ from $\mathsf{NP}$ (in its vanilla form).