= Solution
A finite extension is Galois when it is both normal and separable; over $\mathbb Q$ separability is automatic, so it is equivalently a <splitting field> of a <polynomial> with distinct roots. Its <Galois group> consists of the <field> <automorphisms> fixing $\mathbb Q$. If $G=\operatorname{Gal}(L/\mathbb Q)$, the <Fundamental theorem of Galois theory> gives mutually inverse inclusion-reversing bijections
$$
H\longmapsto L^H,\qquad K\longmapsto\operatorname{Gal}(L/K).
$$
They satisfy $[L:L^H]=|H|$ and $[L^H:\mathbb Q]=[G:H]$. The intermediate extension $K/\mathbb Q$ is Galois exactly when $H=\operatorname{Gal}(L/K)$ is normal, and then $\boxed{\operatorname{Gal}(K/\mathbb Q)\cong G/H}$ via restriction.
Every <permutation> of the four roots permutes the three unordered partitions into two pairs and hence permutes the three $\beta_j$. Thus each coefficient of $g(X)=\prod_{j=1}^3(X-\beta_j)$ is a <symmetric polynomial> in the four roots. The <Fundamental theorem of symmetric polynomials> expresses it in the <elementary symmetric functions>, which are rational coefficients of $f$. Hence $g\in\mathbb Q[X]$. Each product $\alpha_i\alpha_j$, $i<j$, occurs exactly twice in $\beta_1+\beta_2+\beta_3$, so that sum is $2q$. Therefore the quadratic coefficient is $\boxed{-2q}$.
For the given cubic write $g(X)=(X-2)^3+12$. The translated cubic $Y^3+12$ is Eisenstein at three, hence irreducible. It has one real root and a nonreal conjugate pair. Its <Galois group> is transitive on three roots, so its order is divisible by three; <complex conjugation> is a <transposition>. A <subgroup> of $S_3$ with both a three-cycle and a <transposition> is all of $S_3$. Hence
$$
\boxed{\operatorname{Gal}(K/\mathbb Q)\cong S_3,\qquad[K:\mathbb Q]=6.}
$$
Let $G=\operatorname{Gal}(L/\mathbb Q)\le S_4$. Irreducibility of the quartic makes the root action transitive, so four divides $|G|$. The cubic <splitting field> $K$ lies inside $L$ by the definitions of the $\beta_j$, and restriction onto its <Galois group> gives a quotient of order six. Thus six also divides $|G|$. Since $|G|$ divides 24, it is 12 or 24. If it were 12, the supplied index-two result would give $G=A_4$. The restriction kernel would then be a <normal subgroup> of $A_4$ of index six, contrary to the other supplied result. Thus
$$
\boxed{G=S_4.}
$$
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