= Solution
Normalize each finite-place <valuation> to have image $\mathbb Z$. For a radical class $[a]$, the residue of $v(a)$ modulo $n$ is well-defined, because replacing $a$ by $ab^n$ adds $nv(b)$. A suitable finite set is
$$
\boxed{S=S_\infty\ \cup\ \{v<\infty:v(n)>0\}
\ \cup\ \{v<\infty:v(a)\not\equiv0\pmod n
\text{ for some }[a]\in\Delta\}.}
$$
One can compute the last union from any finite generating set of $\Delta$. Each representative has only finitely many nonzero <valuations>, so this indeed defines a finite set depending only on $n$ and the classes, not on a choice of representatives. Including all infinite places safely allows possible real-to-complex ramification.
Fix $v\notin S$ and work over the completion $K_v$. A representative has $v(a)=nk$, so $a=\pi^{nk}u$ with $u$ a unit. Adjoining an $n$th root of $a$ is the same as adjoining an $n$th root of $u$. Since $n$ is invertible in the <residue field>, $X^n-\bar u$ is separable: its derivative $nX^{n-1}$ has no common zero with it.
Take a finite extension of the finite <residue field> containing all roots of all these finitely many reduced polynomials. The corresponding finite <unramified extension> of $K_v$ exists by the classification proved in Question 2. <Hensel's lemma> lifts each simple residue root to a root of the original polynomial in that <unramified extension>. Thus all the required radicals, and every conjugate root, lie in an unramified local extension. Every completion of the radical field at $v$ is therefore unramified. This proves <finite ramification support for radical extensions>.
The local argument does not assume $\mu_n\subset K$: adjoining all roots proves the stronger unramified assertion for the <splitting field>, and hence for any field obtained by choosing the radicals. For $n=1$ the radical extension is trivial; the displayed set is still valid.
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