= Solution
With microstate energies held fixed when differentiating,
$$
\boxed{\langle E\rangle=-\frac{Z'}Z=-\partial_\beta\ln Z,\qquad
\langle E^2\rangle=\frac{Z''}Z,\qquad
\sigma_E^2=\partial_\beta^2\ln Z.}
$$
Since $d\beta/dT=-1/(k_BT^2)$,
$$
\boxed{\sigma_E^2=k_BT^2C_V.}
$$
The <canonical ensemble> fixes <volume> and <particle number>, and any other external <Hamiltonian> parameters. Thus the derivative is $C_V$; pressure-fixed fluctuations would naturally involve <enthalpy> in a different ensemble.
The <translational and two-level energy variance> follows because translation and internal states factorize for this dilute classical gas:
$$
Z=Z_{\rm tr}(1+e^{-\beta\epsilon})^N,\qquad
\ln Z_{\rm tr}=\text{constant}-\frac{3N}{2}\ln\beta.
$$
Differentiating twice yields
$$
\boxed{\sigma_E^2=\frac32N(k_BT)^2+
N\epsilon^2\frac{e^{\epsilon/(k_BT)}}{[1+e^{\epsilon/(k_BT)}]^2}.}
$$
The terms are independent translational and two-level binomial <variances>. At $k_BT\ll\epsilon$, excitation freezes out exponentially and translation dominates. At $k_BT\gg\epsilon$, the internal <variance> tends to $N\epsilon^2/4$ while the translational <variance> grows as $T^2$; the internal <heat capacity> consequently tends to zero. Fractional total-energy fluctuations are of order $N^{-1/2}$ for large $N$. The low-temperature statement presumes translation is still nondegenerate; at strictly zero <temperature> the classical-gas model fails.
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