= Solution
Let $a=\sqrt{2+\sqrt5}>0$ and $b=i/a$, so $b^2=2-\sqrt5$. The roots are $\pm a,\pm b$. The element $2+\sqrt5$ is not a square in $\mathbb Q(\sqrt5)$: its <field norm> is $-1$, whereas the <field norm> of a square is a square of a rational number and hence nonnegative. Thus $[\mathbb Q(a):\mathbb Q(\sqrt5)]=2$, and $[\mathbb Q(a):\mathbb Q]=4$. This also proves irreducibility of the quartic.
The <splitting field> is $K=\mathbb Q(a,i)$, since $b=i/a$ and $i=ab$. Because $\mathbb Q(a)\subset\mathbb R$, adjoining $i$ doubles the degree:
$$
\boxed{[K:\mathbb Q]=8}.
$$
The <Galois group> acts faithfully on the four roots and preserves the partition into the two opposite pairs $\{a,-a\}$ and $\{b,-b\}$, possibly interchanging the pairs. The full permutation group preserving this partition has order eight and is the <dihedral group> of a square. Since the <Galois group> has order eight, it is exactly that group. Equivalently its generators can be chosen as $r:a\mapsto b,\ b\mapsto-a$ and complex conjugation $s:a\mapsto a,\ b\mapsto-b$, with $r^4=s^2=1$ and $srs=r^{-1}$.
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