= Solution
The independent identically distributed <central limit theorem> states that for independent identically distributed <random variables> $U_k$ with finite <mean> $\mu$ and positive finite <variance> $\sigma^2$,
$$
\frac{\sum_{k=1}^nU_k-n\mu}{\sigma\sqrt n}\ \xrightarrow{d}\ N(0,1).
$$
For $0<u<1$, <continuity>, strict monotonicity and the endpoint limits of the <cumulative distribution function> give a unique $x=F^{-1}(u)$. Then $P(F(X)\le u)=P(X\le x)=F(x)=u$. Thus $Y=F(X)$ has the <uniform distribution> on $(0,1)$, by the <probability integral transform>.
For $W=-\log(1-Y)$, $P(W\le w)=1-e^{-w}$ when $w\ge0$, and it is zero otherwise. Therefore its <probability density function> is $e^{-w}\chi_{(0,\infty)}(w)$: the <exponential distribution> of rate one. By <integration by parts>, $E[W]=1$ and $E[W^2]=2$, so $\operatorname{Var}(W)=1$.
The variables $W_k=-\log(1-F_k(X_k))$ are independent because they are separate measurable functions of independent <random variables>. The <probability integral transform> makes all of them identically distributed as $\operatorname{Exp}(1)$, even though the $X_k$ need not be identically distributed. The displayed $Z_n$ is exactly $n^{-1/2}\sum_{k=1}^n(W_k-1)$. The <central limit theorem> therefore gives
$$
\boxed{Z_n\xrightarrow{d}N(0,1).}
$$
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