= Solution
Use the <Probabilists' Hermite polynomials>, rather than the differently scaled physicists' convention:
$$
H_r(x)=(-1)^re^{x^2/2}\frac{d^r}{dx^r}e^{-x^2/2}.
$$
Thus $H_2=x^2-1$, $H_3=x^3-3x$, and $H_5=x^5-10x^3+15x$. The <cumulant-generating function> of an exponential variable of mean $\theta$ is $-\log(1-\theta t)$ near zero, so its $r$th <cumulant> is $(r-1)!\theta^r$. Hence the standardized third and fourth cumulants are $\rho_3=2$ and $\rho_4=6$; the fourth is a cumulant, not the fourth raw central moment.
Write $F_n$ for the <cumulative distribution function> of the pivot in part (i). Its <Edgeworth expansion>, valid here because the exponential law is smooth, nonlattice and has exponential moments, is
$$
F_n(t)=\Phi(t)-\phi(t)\left[\frac{H_2(t)}{3\sqrt n}+\frac{H_3(t)}{4n}+\frac{H_5(t)}{18n}\right]+O(n^{-3/2})
$$
at the fixed endpoints $t=\pm z$. The coverage is exactly $F_n(z)-F_n(-z)$. Since $H_2$ and $\phi$ are even, the $n^{-1/2}$ terms cancel; the two odd Hermite terms double. Consequently the <symmetric Edgeworth coverage cancellation> gives
$$
\boxed{\mathbb P_\theta(\theta\text{ lies in the interval})=1-\alpha-\frac{\phi(z)}n\left[\frac12H_3(z)+\frac19H_5(z)\right]+O(n^{-3/2}).}
$$
In particular its <coverage probability> error is $\boxed{O(n^{-1})}$. This is a fixed-confidence-level expansion; it is not a uniform statement for normal cutoffs increasing with $n$.
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