= Solution
For the first <polynomial>, put $\alpha=\sqrt{1+\sqrt{26}}$ and $\beta=5i/\alpha$, so that $\beta^2=1-\sqrt{26}$. The roots are $\pm\alpha,\pm\beta$, and
$$
\boxed{K=\mathbb Q(\alpha,i),\qquad G\cong D_4\text{ of order }8.}
$$
To justify the degree, a rational quadratic factorization would have the form $(x^2+ux+v)(x^2-ux+w)$. The vanishing linear coefficient implies either $u=0$ or $v=w$. The former requires $v,w=-1\pm\sqrt{26}$, not rational, while the latter requires $v^2=-25$, also impossible over $\mathbb Q$. There is no rational root, since its square would have to be $1\pm\sqrt{26}$, so the <polynomial> is irreducible. The real field $\mathbb Q(\alpha)$ has degree four and does not contain $i$, giving degree eight after adjoining it. The group acts transitively on the four roots while preserving their two opposite pairs, so is contained in the square's <dihedral group> of order eight; the degree makes it the full group.
For the second <polynomial>, choose $\alpha=\sqrt3+i\sqrt2$ and $\beta=\sqrt3-i\sqrt2$. Then $\alpha^2=1+2i\sqrt6$, $\beta^2=1-2i\sqrt6$ and $\alpha\beta=5$. Its four roots generate
$$
\boxed{K=\mathbb Q(\sqrt3,\sqrt{-2}),\qquad G\cong C_2\times C_2.}
$$
Indeed $\alpha+\beta=2\sqrt3$ and $\alpha-\beta=2\sqrt{-2}$ recover both generators. The real quadratic field $\mathbb Q(\sqrt3)$ cannot contain $\sqrt{-2}$, so the degree is four. Independent changes of the two square-root signs give all four <automorphisms>.
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