I think $(3)$ is misleading. Since $E_k$ is the set of all sample points in exactly $k$ of the $A_i$ sets, $(3)$ makes no sense at all to me.
For $(3)$ I suggest we let each $E_k$ be a disjoint union:
$$E_k = \bigcup_{i=1}^{\binom{n}{k}} D_{k,i}$$
where each $D_{k,i}$ is the subset of $E_k$ where the $k$ $A_i$'s are a distinct $k$ of the sets $A_1,\ldots,A_n$.
Then, for $(3),\;$ we assume WLOG, for any given $k,i,\;$ that $D_{k,i}$ is contained in $A_1,\ldots,A_k$ (and in no others) and proceed as I think the author intends, with $D_{k,i}$ instead of $E_k$. Note that, by the additivity of probability:
$$P(E_k) = \sum_{i=1}^{\binom{n}{k}} P(D_{k,i}).$$
The idea of $(5)$ is that we have
\begin{eqnarray*}
P\left(\bigcup_{i=1}^{n}A_i\right) &=& \sum_{k=1}^{n} P(E_k) \qquad\qquad\text{by $(1)$} \\
&=& \sum_{k=1}^{n} \sum_{i=1}^{\binom{n}{k}} P(D_{k,i}) \qquad\text{by $(3)$}\qquad\qquad\qquad\text{(*)} \\
\end{eqnarray*}
By $(3),\;$ any $P(D_{k,i})$ appears in the expression $P_1-P_2+P_3-\cdots\pm P_k,\;$ exactly $\left(k-\binom{k}{2}+\binom{k}{3}-\cdots\pm\binom{k}{k}\right)$ times, and this equals $1$ by $(4)$.
So, from $(*),\;$ we can conclude that $P(\cup_{i=1}^{n}A_i) = P_1-P_2+P_3-\cdots\pm P_n,\;$ as required.
Place the saucers on the table in the order: azure, azure, blue, blue, cyan, cyan.
Let's treat cups of the same color as indistinguishable.
There are $\binom{6}{2}$ ways to choose two of the six positions for the azure cups, $\binom{4}{2}$ ways to choose two of the remaining four positions for the blue cups, and $\binom{2}{2}$ ways to choose both of the remaining two positions for the cyan cups. Hence, the cups may be distributed without restriction in
$$\binom{6}{2, 2, 2} = \binom{6}{2}\binom{4}{2}\binom{2}{2} = \binom{6}{2}\binom{4}{2}$$
distinguishable ways.
From these distributions, we must exclude those in which one or more cups is placed on a saucer of the same color.
Let $A_i$ be the event that an azure cup is placed on the $i$th azure saucer; let $B_i$ be the event that a blue cup is placed on the $i$th blue saucer; let $C_i$ be the event that a cyan cup is placed on the $i$th cyan saucer.
$|A_1|$: Since an azure cup is placed on the first azure saucer, there are two blue cups, two cyan cups, and one azure cup left to distribute to the remaining five saucers. They can be distributed in
$$\binom{5}{1, 2, 2} = \binom{5}{1}\binom{4}{2}\binom{2}{2} = \binom{5}{1}\binom{4}{2}$$
distinguishable ways. By symmetry,
$$|A_1| = |A_2| = |B_1| = |B_2| = |C_1| = |C_2|$$
$|A_1 \cap A_2|$: Since both azure cups have been placed on azure saucers, there are two blue and two cups left to distribute to the remaining four saucers. They can be distributed in
$$\binom{4}{2, 2} = \binom{4}{2}\binom{2}{2} = \binom{4}{2}$$
ways. By symmetry,
$$|A_1 \cap A_2| = |B_1 \cap B_2| = |C_1 \cap C_2|$$
$|A_1 \cap B_1|$: Since an azure cup has been placed on the first azure saucer and a blue cup has been placed on the first blue saucer, there are two cyan cups, one azure cup, and one blue cup left to distribute to the remaining four saucers. They can be distributed in
$$\binom{4}{1, 1, 2} = \binom{4}{1}\binom{3}{1}\binom{2}{2} = \binom{4}{1}\binom{3}{1}$$
distinguishable ways. By symmetry,
$$|A_1 \cap B_1| = |A_1 \cap B_2| = |A_1 \cap C_1| = |A_1 \cap C_2| = |A_2 \cap B_1| = |A_2 \cap B_2| = |A_2 \cap C_1| = |A_2 \cap C_2| = |B_1 \cap C_1| = |B_1 \cap C_2| = |B_2 \cap C_1| = |B_2 \cap C_2|$$
$|A_1 \cap A_2 \cap B_1|$: Since both azure cups have been placed on azure saucers and a blue cup has been placed on the first blue saucer, there are two cyan and one blue cup left to distribute to the remaining three saucers. They can be distributed in
$$\binom{3}{1, 2} = \binom{3}{1}\binom{2}{2} = \binom{3}{1}$$
distinguishable ways. By symmetry,
$$|A_1 \cap A_2 \cap B_1| = |A_1 \cap A_2 \cap B_2| = |A_1 \cap A_2 \cap C_1| = |A_1 \cap A_2 \cap C_2| = |A_1 \cap B_1 \cap B_2| = |A_2 \cap B_1 \cap B_2| = |A_1 \cap C_1 \cap C_2| = |A_2 \cap C_1 \cap C_2| = |B_1 \cap B_2 \cap C_1| = |B_1 \cap B_2 \cap C_2| = |B_1 \cap C_1 \cap C_2| = |B_2 \cap C_1 \cap C_2|$$
$|A_1 \cap B_1 \cap C_1|$: Since an azure cup has been placed on the first azure saucer, a blue cup has been placed on the first blue saucer, and a cyan cup has been placed on the first cyan saucer, we have an azure cup, a blue cup, and a cyan cup to distribute to the remaining three saucers. They can be distributed in
$$\binom{3}{1, 1, 1} = \binom{3}{1}\binom{2}{1}\binom{1}{1} = 3!$$
distinguishable ways. By symmetry,
$$|A_1 \cap B_1 \cap C_1| = |A_1 \cap B_1 \cap C_2| = |A_1 \cap B_2 \cap C_1| = |A_1 \cap B_2 \cap C_2| = |A_2 \cap B_1 \cap C_1| = |A_2 \cap B_1 \cap C_2| = |A_2 \cap B_2 \cap C_1| = |A_2 \cap B_2 \cap C_2|$$
$|A_1 \cap A_2 \cap B_1 \cap B_2|$: Since both azure cups have been placed on azure saucers and both blue cups have been placed on blue saucers, there are two cyan cups to distribute to the remaining two saucers. They can be distributed in
$$\binom{2}{2}$$
distinguishable ways. By symmetry,
$$|A_1 \cap A_2 \cap B_1 \cap B_2| = |A_1 \cap A_2 \cap C_1 \cap C_2| = |B_1 \cap B_2 \cap C_1 \cap C_2|$$
$|A_1 \cap A_2 \cap B_1 \cap C_1|$: Since both azure cups have been placed on azure saucers, a blue cup has been placed on the first blue saucer, and a cyan cup has been placed on the first cyan saucer, we have one blue cup and one cyan cup to distribute to the remaining two saucers. They can be distributed in
$$\binom{2}{1, 1} = \binom{2}{1}\binom{1}{1} = 2!$$
distinguishable ways. By symmetry,
$$|A_1 \cap A_2 \cap B_1 \cap C_1| = |A_1 \cap A_2 \cap B_1 \cap C_2| = |A_1 \cap A_2 \cap B_2 \cap C_1| = |A_1 \cap A_2 \cap B_2 \cap C_2| = |A_1 \cap B_1 \cap B_2 \cap C_1| = |A_1 \cap B_1 \cap B_2 \cap C_2| = |A_2 \cap B_1 \cap B_2 \cap C_1| = |A_2 \cap B_1 \cap B_2 \cap C_2| = |A_1 \cap B_1 \cap C_1 \cap C_2| = |A_1 \cap B_2 \cap C_1 \cap C_2| = |A_2 \cap B_1 \cap C_1 \cap C_2| = |A_2 \cap B_2 \cap C_1 \cap C_2|$$
$|A_1 \cap A_2 \cap B_1 \cap B_2 \cap C_1|$: Since both azure cups have been placed on azure saucers, both blue cups have been placed on blue saucers, and a cyan cup has been placed on the first azure saucer, the remaining cyan cup must be placed on the remaining saucer. There is one way to do this. By symmetry,
$$|A_1 \cap A_2 \cap B_1 \cap B_2 \cap C_1| = |A_1 \cap A_2 \cap B_1 \cap B_2 \cap C_2| = |A_1 \cap A_2 \cap B_1 \cap C_1 \cap C_2| = |A_1 \cap A_2 \cap B_2 \cap C_1 \cap C_2| = |A_1 \cap B_1 \cap B_2 \cap C_1 \cap C_2| = |A_2 \cap B_1 \cap B_2 \cap C_1 \cap C_2|$$
$|A_1 \cap A_2 \cap B_1 \cup B_2 \cap C_1 \cap C_2|$: Since both azure cups have been placed on azure saucers, both blue cups have been placed on blue saucers, and both cyan cups have been placed on cyan saucers, there are no cups left to distribute. There is one way to do this.
Thus, by the Inclusion-Exclusion Principle, the number of ways to distribute the cups so that no cup is on a saucer of the same color is
$$\binom{6}{2}\binom{4}{2} - 6\binom{5}{1}\binom{4}{2} + 3\binom{4}{2} + 12\binom{4}{1}\binom{3}{1} - 12\binom{3}{1} - 8 \cdot 3! + 3 \cdot 1 + 12 \cdot 2! - 6 \cdot 1 + 1$$
Hence, the probability no cup is on a saucer of the same color is
$$\frac{\binom{6}{2}\binom{4}{2} - 6\binom{5}{1}\binom{4}{2} + 3\binom{4}{2} + 12\binom{4}{1}\binom{3}{1} - 12\binom{3}{1} - 8 \cdot 3! + 3 \cdot 1 + 12 \cdot 2! - 6 \cdot 1 + 1}{\binom{6}{2}\binom{4}{2}}$$
Best Answer
In general, if $A_1, A_2,\ldots, A_n$ are mutually disjoint events, then $$ P\Bigl(\,\bigcup\limits_{i=1}^n A_i\,\Bigr ) =\sum_{i=1}^n P(A_i). $$
But (1) does not hold for arbitrary unions. One strategy to find a formula for the probability of a union of arbitrary events is to write the given union as a (different) union of mutually disjoint events.
From this, it follows that your initial statement, "This is the probability of sample points belonging to exactly 1 event, exactly 2 events, ...,exactly $n$ events." , is correct.
$P\Bigl(\,\bigcup\limits_{i=1}^n A_i\,\Bigr )$ is the sum of the probabilities of the events $B_i$ where $B_i$ is the event that exactly $i$ of the events $A_j$ occurs. You have, noting that the $B_i$ are mutually disjoint: $$ P\Bigl(\,\bigcup\limits_{i=1}^n A_i\,\Bigr ) =\sum_{i=1}^n P(B_i). $$
But, the subsequent formula you have is incorrect. In fact, the event $B_1$ is somewhat complicated. If exactly one event $A_i$ occurs, then either $A_1$ only occurs, or $A_2$ only occurs, ... . So $$ B_1= \bigcup_{j=1}^n \Bigl(\,A_j\cap ({\textstyle\bigcup\limits_{i\ne j }}\, A_i^C\,)\,\Bigr). $$
Trying to write down descriptions of the other $B_i$ might lead you to suspect this particular method is more trouble than its worth (of course, it may prove useful in certain problems).
The formula in Dilip's comment is a nice way to "disjointify" the union $\cup A_i$. If you draw a Venn diagram for the case $n=2$, it is easy to convince yourself that $A_1\cup A_2$ can be written as a union of pairwise disjoint events: $$ A_1\cup A_2 = A_1 \cup (A_1^c\cap A_2). $$ So you can write $$ P(A_1\cup A_2) =P( A_1)+ P (A_1^c\cap A_2). $$
For the case $n=3$, $A_1\cup A_2\cup A_3$ can be written as a union of pairwise disjoint events: $$ A_1\cup A_2\cup A_3 = A_1 \cup(A_1^c\cap A_2)\cup (A_1^c\cap A_2^c\cap A_3). $$ So you can write $$ P(A_1\cup A_2\cup A_3) =P( A_1)+P(A_1^c\cap A_2)+P (A_1^c\cap A_2^c\cap A_3). $$
And, of course the general formula in Dilip's comment holds because the union of the $A_i$ is written as a union of pairwise disjoint sets.
The "add and subtract business" you mention seems like what is done in deriving the so-called Inclusion-Exclusion principle, as mentioned in gnometorule's answer. This gives a different method for evaluating the probability of a union (it is not exactly the "disjointification" method above).
As a warm up to the general formula, let's consider the formulas for $P(A\cup B)$ and $P(A\cup B\cup C)$.
Probability of the union of two events:
Let us find the probability of the union of two arbitrary events $A$ and $B$.
One might think $P(A\cup B)=P(A)+P(B)$; however, each of $P(A)$ and $P(B)$ counts the probability of $A\cap B$. We thus have to subtract this probability from $P(A)+P(B)$ to obtain the correct formula:
$$ P(A\cup B)=P(A)+P(B)-P(A\cap B). $$ Another way to derive the formula is to note that $A\cup B$ is the disjoint union of $A/B$, $B/A$, and $A\cap B$. Now, since $A$ is the disjoint union of $A\cap B$ and $A/B$, it follows that $P(A)=P(A\cap B)+P(A/B)$; whence $P(A/B)= P(A)-P(A\cap B)$.
In a similar manner, one shows that $P(B/A)= P(B)-P(B\cap A)$.
It follows that $$\eqalign{ P(A\cup B)&=P(A/B)+P(B/A)+P(A\cap B)\cr &=\bigl( P(A)-P(A\cap B)\bigr)+\bigl(P(B)-P(B\cap A)\bigr)+P(A\cap B) \cr &= P(A)+P(B)-P(A\cap B). } $$
Probability of the union of three events: For three arbitrary events $A $, $B$, and $C$, to find $P(A \cup B\cup C)$, we first start with the guess $$ P(A\cup B\cup C)=P(A )+P(B)+P(C).\tag{1} $$ Now (1) counts each of $P(A \cap B)$, $P(A\cap C)$, and $P(B\cap C)$ twice. To make up for this, we have to subtract $$P(A \cap B)+P(A \cap C)+P(B\cap B).\tag{2}$$ But $P(A )+P(B)+P(C)$ counts $P(A \cap B \cap C)$ thrice, and in (2) we subtracted it thrice. So, we must add $P(A )+P(B)+P(C)$ back in again. We have: $$ P(A\cup B\cup C)=P(A)+P(B)+P(C)-P(A\cap B)-P(A\cap C)-P(B\cap C)+P(A\cap B\cap C). $$ Looking at the Venn diagram may be helpful here:
More generally, for the events $A_1$, $A_2$, $\ldots\,$, $A_n$, we have the inclusion-exclusion principle: $$\eqalign{ P\Bigl(\bigcup_{i=1}^n A_i\Bigr) = \sum_{i\le n} P(A_i) - &\sum_{i_1<i_2} P(A_{i_1}\cap A_{i_2}) +\sum_{i_1<i_2<i_3} P(A_{i_1}\cap A_{i_2}\cap A_{i_3}) - \cr &\cdots+ (-1)^{n}\sum_{i_1<i_2<\cdots<i_{n-1} } P(A_{i_1}\cap\cdots\cap A_{i_{n-1}} )\cr&\qquad\qquad + (-1)^{n+1}P(A_1\cap A_2\cap\cdots\cap A_n)}. $$
As pointed out in the comments, this can be proved by induction.