The events are not mutually exclusive, but I'm assuming for this task, you are to take the events as independent.
For example:
For the first, it suffices to write $A_1 \cup A_2 \cup A_3$. For the second, we have intersection $A_1 \cap A_2\cap A_3$. Do you understand why?
"Or" is usually taken to be "inclusive": denoted by union. $A$ or $B$ or both $A$ and $B$.
"And" is usually taken to be intersection.
If "exclusive or" is specified, so that we have $A$ or $B$ but not both $A$ and $B$, you need to use both union and intersection into account (for example, in $c$).
Then compare how $(c)$ and $(e)$ differ: $(c)$ is a bit more complicated: $(e)$ is simply asserting an inclusive or: $$(e)\quad A_1 \cup (A_2 \cap A_3)$$
whereas $(c)$ is exclusive, so we need to say $$(c)\quad \Big( A_1 \cap (A_2 \cap A_3)'\Big) \cup \Big(A_1' \cap (A_2 \cap A_3)\Big)$$
Can you go back and try to complete tasks? In particular, see if you can write $(d)$ using the given set operations.
With respect to Venn Diagrams, you only need paper and pencil to draw Venn Diagrams. Google "Venn Diagrams" and you'll find some "on-line" software where you can experiment with drawing the appropriate Venn Diagrams.
$A1= A\cap B' \cap C'$ and $A2= A'\cap B \cap C'$ are mutually exclusive since for $A \cap B' \cap C'$ to happen we need $A$ to happen, and for the second one we need $A$ to not happen. So, $A1$ and $A2$ are mutually exclusive. Likewise $A2$ and $A3$ are mutually exclusive, and so are $A1$ and $A3$. So, they are all mutually exclusive.
This can be generalized as follows:
Say we have a numbers of events $B_1$ through $B_n$. Let $A_i$ be the event that $B_i$ happens but none of $B_j$ with $j \not = i$. Then any $A_i$ and $A_j$ with $i \not = j$ are mutually exclusive, since for $A_i$ to happen you need $B_i$ to happen, but for $A_j$ to happen you need $B_j$ to not happen. So, all the $A_i$'s are mutually exclusive.
Best Answer
One of the axioms of probability is that if $A_1, A_2, \dots$ are disjoint, then
$$\begin{align} \mathbb{P}\left(\bigcup_{i=1}^{\infty}A_i\right) = \sum\limits_{i=1}^{\infty}\mathbb{P}\left(A_i\right)\text{.}\tag{*} \end{align}$$
It so happens that this is also true if you have a finite number of disjoint events. If you're interested in more detail, consult a measure-theoretic probability textbook.
Let's motivate the proof for the probability of the union of three events by using this axiom to prove the probability of the union of two events.
Now, armed with the result that we proved in the previous theorem, we can now prove the result for the probability of the union of three events.