= Solution
For $L=\mathbb Q_p(\zeta_p)$, put $\pi=\zeta_p-1$. The <polynomial>
$$
\Phi_p(1+T)=\frac{(1+T)^p-1}{T}
$$
is <Eisenstein> at $p$. It gives $[L:\mathbb Q_p]=e=p-1$, total ramification, and $v_L(\pi)=1$. Also $\mathcal O_L=\mathbb Z_p[\pi]$: in the power <basis>, terms $c_i\pi^i$ have <valuations> $(p-1)v_p(c_i)+i$ in different residue classes modulo $p-1$, so cannot cancel at the minimum. Integrality forces all $c_i$ integral.
Use part (b) with $\Phi_p(T)$ and differentiate the quotient at $\zeta_p$:
$$
\Phi'_p(\zeta_p)=\frac{p\zeta_p^{p-1}}{\zeta_p-1}.
$$
The <root of unity> is a <unit>, while $v_L(p)=p-1$ and $v_L(\pi)=1$. Hence the <cyclotomic local different exponent> is
$$
\boxed{\delta(\mathbb Q_p(\zeta_p)/\mathbb Q_p)=p-2.}
$$
This includes $p=2$: that extension is trivial and the exponent is zero.
For $m=p^2-1$, construct an unramified quadratic extension $E/\mathbb Q_p$ by lifting an irreducible <quadratic polynomial> over $\mathbb F_p$. Its <residue field> is $\mathbb F_{p^2}$. Each of its nonzero residue elements is a simple <polynomial root> of $T^m-1$, so <Hensel's lemma> lifts all $m$ <polynomial roots> into $E$. A primitive residue element lifts to a primitive mth <polynomial root>: its reduction already has order $m$, and its mth power is one.
Such a primitive <polynomial root> cannot generate a proper subfield of $E$, since its residue has order $p^2-1>p-1$ and therefore generates the quadratic residue extension. Thus $\mathbb Q_p(\zeta_m)=E$ is unramified of degree two. Part (a) gives $\mathcal O_E=\mathbb Z_p[\zeta_m]$. The reduced <minimal polynomial of an algebraic element> has degree two and is separable, so its derivative at $\zeta_m$ is a <unit>. Part (b) gives
$$
\boxed{\delta(\mathbb Q_p(\zeta_{p^2-1})/\mathbb Q_p)=0.}
$$
Both answers use normalized <discrete valuations> on the top <field>.
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