= Solution
Put $a=\sqrt[3]{2}$ and $\zeta=\zeta_3$. The polynomial $X^3-2$ is irreducible over $\mathbb Q$ by the <Eisenstein criterion> at $2$, so $[\mathbb Q(a):\mathbb Q]=3$. This field is real and cannot contain the nonreal $\zeta$. Adjoining $\zeta$ gives a quadratic extension and hence $[F:\mathbb Q]=6$. Since $F$ is the <splitting field> of $X^3-2$, its <Galois group> acts faithfully on $a,\zeta a,\zeta^2a$ and has order six. Consequently
$$
\operatorname{Gal}(F/\mathbb Q)\simeq S_3.
$$
For a prime $\mathfrak q$ over a rational prime $p$, its <absolute residue degree> is the order of the cyclic <quotient group> $D_{\mathfrak q}/I_{\mathfrak q}$ from part (i). If that degree were six, then $D_{\mathfrak q}$, a subgroup of $S_3$, would have order at least six, so $D_{\mathfrak q}=S_3$ and $I_{\mathfrak q}=1$. This would make $S_3$ cyclic, a contradiction. The argument applies also at ramified primes. Thus
$$
\boxed{\text{No prime of }F\text{ has absolute residue degree }6.}
$$
For the second request printed with the same subpart label, work over $E=\mathbb Q(\zeta)$. The <Galois group> of $F/E$ is cyclic of order three, with generator $\sigma(a)=\zeta a$ and $\sigma(\zeta)=\zeta$. The <ideal norm> of $(\zeta+3)$ is seven, since
$$
(\zeta+3)(\zeta^2+3)=7.
$$
In its <residue field> we have $\zeta\equiv-3\equiv4\pmod{\mathfrak p_1}$. The polynomial $X^3-2$ has no root in $\mathbb F_7$, because its cube residues are $0,1,6$. It is therefore irreducible there. Its discriminant $-108$ is a unit at $\mathfrak p_1$. To check the local index condition, the <trace pairing> matrix of $1,a,a^2$ has this unit determinant. For any integral element $b$ of $F$, its traces against these three basis elements are integral; solving with this invertible matrix shows that its basis coefficients are integral locally. Thus the relative <ring of integers> equals $\mathcal O_E[a]$ at $\mathfrak p_1$. The relative <Kummer-Dedekind theorem> gives one unramified prime above $\mathfrak p_1$, of <residue degree> three.
Use <arithmetic Frobenius>, characterized by $b\mapsto b^7$ on this <residue field>. As $a^3=2$,
$$
a^7=4a\equiv\zeta a\pmod{\mathfrak q}.
$$
The three possible images $a,\zeta a,\zeta^2a$ remain distinct modulo $\mathfrak q$, since $a$ is nonzero and $1,4,2$ are distinct modulo seven. This congruence therefore identifies the actual automorphism uniquely:
$$
\boxed{\operatorname{Frob}_{\mathfrak p_1}(a)=\zeta_3a,\qquad
\operatorname{Frob}_{\mathfrak p_1}(\zeta_3)=\zeta_3.}
$$
The geometric Frobenius convention would give its inverse. Since $F/E$ is abelian, the arithmetic element is independent of the choice of prime above $\mathfrak p_1$.
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