"What does 𝑖.𝑜 mean?"
As already mentioned in the comments $i.o.$ stands for infinitely often and so
$$\{A_n\,i.o\}=\{\omega\in\Omega:\omega\in A_n \text{ for an infinite number of indices } n\in\{1,2,\dots\}\}.$$
For a more intuitive explanation see the following example:
Consider the probability model for Bernoulli sequences (coin flips). Let $A$ be the event where the pattern $THTH$ occurs infinitely often, i.e., $$A:=\{A_n\,i.o\},$$
where $A_n$ is the event where the pattern $THTH$ occurs beginning at the $n$th coin flip. For example, $$A_2=\{\omega\in\Omega: \eta_2(\omega)=1,\eta_3(\omega)=-1,\eta_4(\omega)=1,\eta_5(\omega)=-1\},$$where we denoted the outcome of the coin flip by $\eta_1,\eta_2,\dots,$ which takes values $+1$ (tails) or $-1$ (heads).
Notice that the set $$\bigcup_{n\ge k} A_n$$ is the set of coin tossing sequences where the pattern occurs starting at some flip $\ge k$. If the pattern only occurs finitely many times in a particular sequence of flips, then that sequence will not be in this set for a large enough $k.$ Therefore, if we take the intersections of all these unions, we get the set of flips where the sequence occurs infinitely many times, i.e., $$A=\{A_n\,i.o\}=\bigcap_{k=1}^\infty\bigcup_{n\ge k} A_n=:\limsup_{n\to\infty} A_n.$$
For the first exercise, if you already proved that $$P\left(\bigcup_{n=1}^\infty C_n\right)=\lim_{n\to\infty}P\left( C_n\right),$$ for expanding sequence of events, i.e., $C_1\subset C_2\subset \cdots,$ you can apply this statement to the complements of $B_n$'s, since $\left(\bigcap_{n=1}^\infty B_n\right)^c=\bigcup_{n=1}^\infty B_n^c$ and $B_n^c$'s are expanding. Using the fact that $P(D^c)=1-P(D)$, we deduce the result.
If you didn't prove the above statement, here is a hint to do so: Write $\bigcup_{n=1}^\infty C_n=C_1\cup C_2\cup \cdots$ as the union of a sequence of disjoint sets: first start with $C_1,$ then add a disjoint set to obtain $C_1\cup C_2$, then again add a disjoint set to obtain $C_1\cup C_2\cup C_3$ and so on. Then use the definition of the probability measure to add those disjoint sets.
For the second exercise, first notice that $$ \bigcup_{n\ge k} A_n=A_k\cup \bigcup_{n\ge k}(A_{n+1}\setminus A_{n})=A_k\cup \bigcup_{n\ge k} (A_n^c\cap A_{n+1}).$$ As you noticed, the sequence $\left\{\bigcup_{n\ge k} A_n\right\}_k$ is a contracting sequence. Therefore, we can use the previous exercise to deduce
$$P\left(\bigcap_{k=1}^\infty \bigcup_{n\ge k} A_n\right) =\lim_{k\to\infty}P\left(\bigcup_{n\ge k} A_n\right)=\lim_{k\to\infty}P\left(A_k\cup \bigcup_{n\ge k} (A_n^c\cap A_{n+1}) \right). $$ Now, by subadditivity,
$$\lim_{k\to\infty}P\left(\bigcup_{n\ge k} A_n\right)\le \lim_{k\to\infty} P(A_k) + \lim_{k\to\infty}\sum_{n\ge k}P\left( A_n^c\cap A_{n+1} \right).$$
The first term on the RHS goes to zero by assumption and the second term also goes to zero by the convergence of the sum $\sum^\infty_{n=1}P(A^c_n\cap A_{n+1}),$ which is also discussed in the comments. Hence, we have that $$\lim_{k\to\infty}P\left(\bigcup_{n\ge k} A_n\right)=P\left(\bigcap_{k=1}^\infty \bigcup_{n\ge k} A_n\right) = P(A_n\,i.o)=0.$$
Best Answer
You may have already caught this error, but the line $$P(\bigcup_{m=n}^{\infty} A_m) = \sum_{m=n}^{\infty} P(A_m) $$ would be correct if the events were disjoint, and independent events are only disjoint in the trivial case where at least one has probability 0.
The ($\Rightarrow$) direction of the proof is trivial; if $\omega$ is such that $\omega \in A_n$ for infinitely many $A_n$, then in particular $\omega \in A_n$ for at least one $A_n$.
For the other direction, fix an arbitrary integer $n$, and consider the decomposition $$\bigcup_{m=1}^{\infty} A_m = \bigcup_{m=1}^n A_m \cup \bigcup_{m=n+1}^{\infty} A_m.$$ You know the probability of the left side is 1, and you can show that the probability of the middle term is strictly less than one by considering its complementary event. You can then use the inclusion-exclusion principle and independence of the two terms on the right to argue that the probability of the rightmost term is one, which is what you need.