= Solution
Write $\alpha=\lambda^{1/5}$, $M=\mathbb Q(\alpha)$ and $E=\mathbb Q(\zeta_5)$. The <Eisenstein criterion> at $\lambda$ gives $[M:\mathbb Q]=5$, whereas the <prime cyclotomic field degree> gives $[E:\mathbb Q]=4$. The <splitting field> $F=E(\alpha)$ is a <Galois extension> of $E$, and an automorphism fixing $E$ sends $\alpha$ to $\zeta_5^b\alpha$. Thus its <Galois group> embeds in $C_5$. It is nontrivial, since $M$ cannot be contained in the degree-four field $E$. Therefore $[F:E]=5$ and $[F:\mathbb Q]=20$.
Index the five roots by $j\in\mathbb F_5$, with root $\zeta_5^j\alpha$. Every automorphism has the form $\zeta_5\mapsto\zeta_5^u$, $\alpha\mapsto\zeta_5^b\alpha$ and acts on indices as $j\mapsto uj+b$, with $u\in\mathbb F_5^\times$. This is a faithful inclusion of $G=\operatorname{Gal}(F/\mathbb Q)$ into the <affine group over a finite field>, and both groups have order $20$, so it is an isomorphism. This is the <affine Galois group of a prime-radical splitting field>.
Let $D$ be the specified <decomposition group>. Its nonidentity element is an <involution>. If its linear coefficient were $1$, it would be a nontrivial translation, which has order $5$. Hence it acts as $j\mapsto-j+b$. There is exactly one fixed root, at $j=b/2$, and the other four roots form two pairs. By <involution orbits in an affine group of odd prime degree>, $D$ has three orbits on the five roots. Those roots identify with $G/\operatorname{Gal}(F/M)$, so the double-coset formula gives three primes in $M$.
For $E$, the embeddings identify with $G/\operatorname{Gal}(F/E)\simeq\mathbb F_5^\times$. The image of $D$ is $\{1,-1\}$ and acts by multiplication, giving the two orbits $\{1,-1\}$ and $\{2,-2\}$. Therefore
$$
\boxed{\#\{\mathfrak m\mid p\text{ in }M\}=3,\qquad\#\{\mathfrak e\mid p\text{ in }E\}=2.}
$$
The orbit argument counts primes whether or not $p$ is ramified; cyclicity of the given order-two <decomposition group> does not require that it be generated by an unramified Frobenius element.
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