= Solution
Let $E=\mathbb Q(\alpha,\beta)$. Since
$$
\sqrt3=\alpha^2-3,\qquad
\sqrt6=\alpha\beta,\qquad
\sqrt2=\frac{\sqrt6}{\sqrt3},
$$
part (b) gives $[E:\mathbb Q]=8$. The roots of $X^4-6X^2+6$ are $\mathord\pm\alpha,\mathord\pm\beta$, all in $E$, so $E$ is its splitting field and is Galois. Its action on the four roots is transitive; the order-eight entry in the list from part (a) is $D_8$. Therefore
$$
\operatorname{Gal}(E/\mathbb Q)\cong D_8.
$$
For an explicit <dihedral Galois action on four radical roots>, define
$$
r(\alpha)=\beta,\qquad r(\beta)=-\alpha,\qquad
s(\alpha)=\alpha,\qquad s(\beta)=-\beta.
$$
Then $r^4=s^2=1$ and $srs=r^{-1}$. The subgroup lattice is determined by
$$
\begin{array}{c}
D_8\\[1mm]
\langle r\rangle\qquad
\langle r^2,s\rangle\qquad
\langle r^2,rs\rangle\\[1mm]
\langle r^2\rangle\quad
\langle s\rangle\quad\langle r^2s\rangle\quad
\langle rs\rangle\quad\langle r^3s\rangle\\[1mm]
\{1\}.
\end{array}
$$
Here $\langle r^2\rangle$ lies in all three order-four subgroups; $\langle s\rangle,\langle r^2s\rangle$ lie in $\langle r^2,s\rangle$; and $\langle rs\rangle,\langle r^3s\rangle$ lie in $\langle r^2,rs\rangle$.
Reversing inclusions under the <Galois correspondence> gives the complete field lattice
$$
\begin{array}{c}
E\\[1mm]
\mathbb Q(\sqrt2,\sqrt3)\quad
\mathbb Q(\alpha)\quad\mathbb Q(\beta)\quad
\mathbb Q(\alpha+\beta)\quad\mathbb Q(\alpha-\beta)\\[1mm]
\mathbb Q(\sqrt2)\qquad
\mathbb Q(\sqrt3)\qquad
\mathbb Q(\sqrt6)\\[1mm]
\mathbb Q.
\end{array}
$$
The incidence is likewise reversed: $\mathbb Q(\sqrt2)$ lies in $\mathbb Q(\sqrt2,\sqrt3)$; $\mathbb Q(\sqrt3)$ lies in that field, $\mathbb Q(\alpha)$, and $\mathbb Q(\beta)$; and $\mathbb Q(\sqrt6)$ lies in that field, $\mathbb Q(\alpha+\beta)$, and $\mathbb Q(\alpha-\beta)$. The hint verifies the last two quartic fields through $\alpha+\beta=\sqrt2\gamma$ and $\alpha-\beta=\sqrt2\delta$.
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