= Solution
We first prove the simple-root <Hensel lemma> directly. Put $x_1=x$. Suppose $x_n\equiv x\pmod\pi$ and $f(x_n)\equiv0\pmod{\pi^n}$. The derivative remains a unit since its reduction agrees with $f'(x)$. Choose $t_n\in\mathcal O_K$ whose residue solves
$$
\frac{f(x_n)}{\pi^n}+t_nf'(x_n)\equiv0\pmod\pi,
$$
and set $x_{n+1}=x_n+\pi^nt_n$. Polynomial expansion gives $f(x_{n+1})=f(x_n)+\pi^nt_nf'(x_n)+\pi^{2n}B$ with $B\in\mathcal O_K$. Since $2n\ge n+1$, the new value vanishes modulo $\pi^{n+1}$. The sequence is <Cauchy> because $v(x_{n+1}-x_n)\ge n$, and <completeness> supplies $y\in\mathcal O_K$ with $y\equiv x\pmod\pi$. Continuity gives $f(y)=0$.
For uniqueness, if $y,z$ are two roots in that residue class, then
$$
0=f(y)-f(z)=(y-z)\bigl(f'(z)+(y-z)B_1\bigr),\qquad B_1\in\mathcal O_K.
$$
The factor in parentheses is a unit, since $f'(z)$ is a unit and $y-z\in\pi\mathcal O_K$. Thus $y=z$. This proves \b[existence and uniqueness of the specified lifted root].
Now suppose the <residue field> has $q=p^r$ elements and put $e=v(p)$. Every element of $k_K^\times$ is a simple root of $T^{q-1}-1$. The <Hensel lemma> therefore lifts its $q-1$ roots uniquely to a subgroup of the <roots of unity> in $K$. These are the <Teichmuller representatives> and give all the prime-to-$p$ torsion. To verify the latter assertion, if a <root of unity> $u\equiv1$ has order $m$ prime to $p$, write $u=1+t$. In $u^m-1=mt+\binom m2t^2+\cdots$, the first term has <valuation> $v(t)$ and all later terms have greater <valuation>, so the sum cannot vanish unless $t=0$. Reduction is therefore injective on prime-to-$p$ torsion, whose order divides $q-1$.
Any <root of unity> has a unique decomposition into a $p$-power-order factor and a prime-to-$p$ factor, by splitting its finite order using coprime powers. For a primitive $p^a$th root $\zeta$, $a\ge1$, reduction in characteristic $p$ gives $\bar\zeta=1$. The cyclotomic identity
$$
p=\Phi_{p^a}(1)=\prod_{\substack{1\le j\le p^a\\p\nmid j}}(1-\zeta^j)
$$
shows that all factors have the same <valuation>: $(\zeta^j-1)/(\zeta-1)=1+\zeta+\cdots+\zeta^{j-1}$ reduces to $j\ne0$. Consequently the <valuation of a primitive p-power root of unity> is
$$
\boxed{v(\zeta-1)=\frac{e}{p^{a-1}(p-1)}.}
$$
The left side is a positive integer. Hence $p^{a-1}(p-1)\le e$, so the possible $p$-power orders are bounded. Combining this bound with the prime-to-$p$ result bounds all root orders by one integer, proving finiteness. A finite multiplicative subgroup of a field is cyclic, and it contains exactly the full prime-to-$p$ subgroup of order $q-1$. Thus
$$
\boxed{|\mu(K)|=p^s(q-1)\quad\text{for some }s\ge0.}
$$
If $e<p-1$, even a primitive $p$th root would violate the <valuation> formula, so \b[$s=0$].
At the boundary $e=p-1$, assume first that $p$ is odd. Take $K_+=\mathbb Q_p(\zeta_p)$. The polynomial $\Phi_p(1+T)=p+\binom p2T+\cdots+T^{p-1}$ is Eisenstein, so $K_+/\mathbb Q_p$ is totally ramified of degree $p-1$ and $v_{K_+}(p)=p-1$. It contains a primitive $p$th root, while the <valuation> formula excludes primitive $p^2$th roots. Thus $s=1$.
For the other example, take $K_0=\mathbb Q_p(\pi)$ with $\pi^{p-1}=p$. The polynomial $T^{p-1}-p$ is Eisenstein, giving the same ramification index and <residue field> $\mathbb F_p$. If a primitive $p$th root belonged to $K_0$, the <valuation> formula would give $\zeta_p-1=u\pi$ with $u$ a unit. Substitute this into $\Phi_p(1+(\zeta_p-1))=0$ and divide by $p$. All intermediate terms reduce to zero, leaving $1+\bar u^{p-1}=0$ in $\mathbb F_p$. But $\bar u^{p-1}=1$, contradicting $2\ne0$. Thus $s=0$ in $K_0$. This proves both cases of <roots of unity at the tame ramification boundary> for odd primes.
\b[The printed boundary claim needs the qualification $p>2$.] When $p=2$, every characteristic-zero field already contains the root $-1$ of order two, so $s=0$ is impossible. If $v(2)=1$, the <valuation> formula excludes order four and higher, forcing $s=1$. For example $K=\mathbb Q_2$ has exactly this behavior. This is a genuine exception to the final request as printed, rather than an omitted construction.
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