The equation where you've enclosed the first part in red is the definition of the $\bigcup_{n=1}^{\infty} S_n$ notation.
It works just like sum notation does:
$$ \sum_{n=a}^{b} f(n) \quad\text{means}\quad f(a)+f(a+1)+\cdots+f(b-1)+f(b) $$
and
$$ \bigcup_{n=a}^{b} f(n) \quad\text{means}\quad f(a)\cup f(a+1)\cup\cdots\cup f(b-1)\cup f(b) $$
$$ \bigcap_{n=a}^{b} f(n) \quad\text{means}\quad f(a)\cap f(a+1)\cap\cdots\cap f(b-1)\cap f(b) $$
When the upper limit is $\infty$ it means a union of infinitely many sets:
$$ \bigcup_{n=a}^{\infty} f(n) \quad\text{means}\quad f(a)\cup f(a+1)\cup\cdots $$
whose precise meaning is defined in the explanation you quote.
The properties that you mention in your question are traditionally referred to as "continuity from above" and "continuity from below" respectively (see the math.stackexchange question Continuity from below and above for a justification of this terminology).
As for understanding how these properties follow from $\sigma$-field axioms such as
Countable additivity: if $\{E_n:n\in\mathbb N\}$ are disjoint events,
then $$\Pr\left[\bigcup_nE_n\right]=\sum_n\Pr[E_n]=\lim_{n\to\infty}\sum_{i=1}^n\Pr[E_i],$$
one of the best ways to get an idea of how to proceed is to look at drawings of sets.
Let's look at the example where $A_1\subset A_2\subset A_3\subset A_4\subset\cdots$
(continuity from below).
One can represent these sets graphically as (only $A_1,A_2,A_3$ are drawn):
Suppose that we define sets $B_1,B_2,B_3,\ldots$ as
$$B_1=A_1\\
B_2=A_2\setminus A_1\\
B_3=A_3\setminus(A_1\cup A_2)\\
\ldots\\
B_n=A_n\setminus(A_1\cup\cdots\cup A_{n-1})\\
\ldots,$$
and so on.
Graphically,
this would look like:
Then,
it seems intuitive enough that the union of every $A_i$ up to $n$ and the union of every $B_i$ up to $n$ are the same for every $n$,
and that the $B_i$ are disjoint (though one still needs to demonstrate this rigorously).
With this in mind,
the solution should become clear (hint: use countable additivity).
Using similar reasoning by drawings/illustrations,
you should be able to solve your second problem.
(Note: I did not create the illustrations in this post.)
Best Answer
I denote the complement of a set $E$ by $E^c$.
Because $P(A_k)+P(A_k^c)=1$ and $P(A) + P(A^c) = 1$, it suffices to show that $\lim_{k \to \infty} P(A_k^c) = P(A^c)$.
Note that $A_1^c \subseteq A_2^c \subseteq A_3^c \subseteq \cdots$. Let $B_k = A_k^c \setminus A_{k-1}^c$. Then the $B_k$ are disjoint and $\bigcup_{k=1}^\infty B_k = \bigcup_{k=1}^\infty A_k^c$. Using axiom (1), you have $$P(A^c) = P(\bigcup_{k=1}^\infty A^c_k) = P(\bigcup_{k=1}^\infty B_k) = \sum_{k=1}^\infty P(B_k) = \sum_{k=1}^\infty (P(A_k^c) - P(A_{k-1}^c)) = \lim_{k \to \infty} P(A_k^c). $$