= Solution
For $B_n=\{X_n>a_n\}$, the definition of the common <cumulative distribution function> gives
$$
\mathbb P(B_n)=1-F(a_n).
$$
The strict exceedance matters here: an <atom of a measure> at $a_n$ belongs to $\{X_n\leq a_n\}$, so it is already included in $F(a_n)$. The <events> $B_n$ are mutually <independent events>, since each depends on a separate member of the <independent random variables> $X_n$.
If $\sum_n(1-F(a_n))<\infty$, <countable subadditivity> gives, for every $N$,
$$
\mathbb P\left(\bigcup_{n\geq N}B_n\right)
\leq\sum_{n\geq N}(1-F(a_n))\longrightarrow0.
$$
<Continuity of probability> then gives $\mathbb P(\limsup_nB_n)=0$. This is the <First Borel-Cantelli lemma> and does not need <independence>.
If the <series> diverges, <independence> and $1-u\leq e^{-u}$ give, for every fixed $N$,
$$
\mathbb P\left(\bigcap_{n=N}^{M}B_n^c\right)
=\prod_{n=N}^{M}F(a_n)
\leq\exp\left(-\sum_{n=N}^{M}(1-F(a_n))\right)\longrightarrow0.
$$
Taking $M\to\infty$ shows that the <probability> of no exceedance after $N$ is zero. A <countable union> over $N$ shows that finitely many exceedances has <probability> zero. This is the <Second Borel-Cantelli lemma>. Together with the preceding <event> equality, it proves
$$
\boxed{\mathbb P(A)=
\begin{cases}
0,&\sum_{n\geq1}(1-F(a_n))<\infty,\\
1,&\sum_{n\geq1}(1-F(a_n))=\infty.
\end{cases}}
$$
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