\begin{proof} PROVING (i): By Extension Theorem we have that $$\begin{align*} \mathbb{P}(A_{n}\text{ i.o.})=\mathbb{P}\left(\limsup_{ n \to \infty }A_{n} \right)&= \mathbb{P}\left( \bigcap_{n=1}^{\infty}\bigcup_{k=n}^{\infty}A_{k} \right) &\text{by definition}\\ &\le \mathbb{P}\left( \bigcup_{k=n}^{\infty}A_{k} \right)&\text{monotonicity}\\ &\le \sum_{k=n}^{\infty}\mathbb{P}(A_{k})&\text{countable subadditivity} \end{align*}$$Provided that the last last sum is finite, then as $n\to \infty$ $\mathbb{P}\left( \limsup_{ n \to \infty } A_{n}\right)=0$ which gives us what we want.

PROVING (ii): So here we work with the complement of the event: $$\{ A_{m}\text{ i.o.} \}^{c}=\{ A_{m}\text{ finitely often} \}\subseteq \bigcup_{N\ge 1}\{ A_{N},A_{N+1},\dots \text{ don't occur} \}$$and by Probability Measure we have $$\mathbb{P}(\{ A_{m}\text{ i.o.} \}^{c})\le \sum_{N=1}^{\infty}\mathbb{P}(\{ A_{N},A_{N+1},\dots \text{ don't occur} \})$$but, we have that $$\begin{align*} \mathbb{P}(\{ A_{N},A_{N+1},\dots \text{ don't occur} \})&= \prod_{n=N}^{\infty}(1-\mathbb{P}(A_{n}))\\ &\le \prod_{n=N}^{\infty}e^{-\mathbb{P}(A_{n})}&\text{since }1-x\le e^{-x}\\ &= e^{-\sum_{n=N}^{\infty}\mathbb{P}(A_{n})}\\ &= e^{-\infty}=0 \end{align*}$$Hence we get that $\forall N:\mathbb{P}(\{ A_{N},A_{N+1},\dots \text{ don't occur} \})=0$ giving us that $\mathbb{P}(\{ A_{m}\text{ i.o.} \}^{c})=0$ and hence $$\mathbb{P}(\{ A_{m}\text{ i.o.} \})=1$$ \end{proof} >[!rmk] >Note that independence is required for the second condition. A counterexample is for infinite coin tosses, let $A_{1}=A_{2}=\dots=$ s.t.