= Solution
The <central limit theorem> states that for independent identically distributed variables of finite mean $\mu$ and positive finite <variance> $\sigma^2$,
$$
\frac{\sum_{j=1}^nX_j-n\mu}{\sigma\sqrt n}\xrightarrow d N(0,1).
$$
For a continuous strictly increasing <cumulative distribution function>, its inverse is defined on $(0,1)$. Thus $\mathbb P(F(X)\le y)=\mathbb P(X\le F^{-1}(y))=y$ for $0<y<1$, proving the uniform law.
If $T=-\log(1-Y)$, then for $t\ge0$, $\mathbb P(T\le t)=\mathbb P(Y\le1-e^{-t})=1-e^{-t}$. Hence $f_T(t)=e^{-t}\mathbf1_{(0,\infty)}(t)$, an exponential density of rate one. Its first two moments are one and two, by <integration by parts>, so $\mathbb ET=1$ and $\operatorname{Var}T=1$.
Although the original variables may have different distributions, their separate transforms $Y_k=F_k(X_k)$ are independent and all uniform. Consequently $T_k=-\log(1-Y_k)$ are independent identically distributed exponentials of rate one, and the statistic is exactly
$$
Z_n=\frac{\sum_{k=1}^nT_k-n}{\sqrt n}.
$$
The stated <central limit theorem> therefore yields $\boxed{Z_n\xrightarrow d N(0,1)}$.
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