= Solution
A primitive tenth root $\xi$ has $\xi^5=-1$ and $\xi\ne-1$, so division of $X^5+1$ by $X+1$ gives
$$
\Phi_{10}(X)=X^4-X^3+X^2-X+1.
$$
It is irreducible over $\mathbb Q$: replacing $X$ by $-X$ gives $\Phi_5(X)$, and then replacing $X$ by $X+1$ gives $X^4+5X^3+10X^2+10X+5$. Any nontrivial monic integer factorization would reduce modulo five to factors that are powers of $X$, forcing both constant terms divisible by five and their product divisible by $25$, contrary to the constant term five. Rational factorization reduces to integer factorization by <Gauss lemma>: products of primitive integer <polynomials> are primitive, since reduction modulo any prime preserves a nonzero product. Thus
$$
\boxed{\operatorname{min}_{\mathbb Q}(\xi)=X^4-X^3+X^2-X+1.}
$$
Let $t=\xi+\xi^{-1}$. Dividing $\Phi_{10}(\xi)=0$ by $\xi^2$ gives $t^2-t-1=0$. Hence $(2t-1)^2=5$, so $\boxed{\sqrt5\in\mathbb Q(\xi)}$. Either sign of this square root suffices; for $\xi=e^{\pi i/5}$ the positive one is $2t-1$.
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