I report here the Exercise with my sketch of a solution.
(a) By Exercise 30 Chapter 3, I know that there are infinitely many primes $q \equiv 1 \bmod r$. Moreover, I know that the ring of integers of $K$ is $S=\mathbb{Z}[\zeta_q]$ and $K = \mathbb{Q}[\zeta_q]$. Using Dedekind's Theorem (Theorem 27 in Chapter 3) I know that, if the cyclotomic polynomial $\Phi_q(x) \in \mathbb{F}_p$ is factored into $r$ irreducible monic polynomials, then also $p$ does in the ring of integers $S$. By properties of cyclotomic polynomials, it's enough taking $p$ such that $p^{(q-1)/r} \equiv 1 \bmod q$ (of course there are infinitely many of such $p$'s).
(b) If $p^{(q-1)/r} \equiv 1 \bmod q$ with $r \neq 1$, then $\Phi_q(x) \in \mathbb{F}_p$ splits into $r$ monic irreducible polynomials of degree $(q-1)/r$. Using again Theorem 27, we know that the inertial degree is $f(Q_i|p) = (q-1)/r$ where $Q_i$ are those prime ideals arising from the factorization of $\Phi_q(x)$. Then, it's enough picking $q \equiv 1 \bmod rf$ to satisfy this point.
(c) There are infinitely many primes $p \equiv 1 \bmod e$. Then? I'm stuck
(d) No ideas
(e) I think $p=5$ and $q=31$ could work ($p^3 \equiv 1 \bmod 31$ and $p \equiv 1 \bmod 2$, moreover $q \equiv 1 \bmod 5$ and $q \equiv 1 \bmod 15$).
What do you think about the first two points and the latter one? Do they work?
Can you give me some hints for point (c) and (d) please?
( Here, Theorem 1 is Theorem 27 in Marcus'book, and here there is a similar solution for point (a))
(c) In the first point you can choose q s.t is equal to 1 module ref.
Then, when you choose p you know that is equal to a module q, as you can see here (Marcus Number Fields Chapter 4 Exercise 8), and that p is equal to 1 module e.
So you have two different conditions on p, one about his residue module q and one about his residue module e.
Since e and q are coprime this two conditions are just one condition module e*q thanks to the Chinese Reminder Theorem.
Such a prime p exists thanks to Dirichlet's theorem.