For the first problem, you did not quite use inclusion/exclusion. Let us count the number that contain the substring $121$. From your $6\cdot 9^5$ we need to subtract the sequences that have been double-counted by $6\cdot 9^5$.
How many sequences are there that have two or more substrings of shape $121$? It is trickier than it looks. We have counted $121121xy$ twice, but we also have counted $12121xyz$ twice.
After you do the subtraction, it will turn out you have subtracted too much, for example the sequences $1212121x$, so they will have to be added back.
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.
Best Answer
You have $r$ properties, denoted by $a_1,a_2,\dots,a_r$. For any set $S\subseteq\{a_1,\dots,a_r\}$ of properties, $N(S)$ is the number of objects having all of the properties in $S$. For example, $N\big(\{a_1\}\big)$ is the number of objects with property $a_1$, and $N\big(\{a_1,a_2\}\big)$ is the number with both property $a_1$ and property $a_2$. (This notation is slightly different from the notation in the question, but it makes things a little easier.) For $i=0,\dots,r$, $e_i$ is the number of objects with exactly $i$ of the $r$ properties, and $s_i$ is the number with at least $i$ of the properties. The total number of objects is $n$; clearly every object has at least $0$ of the properties, so $s_0=n$.
It’s easy to express the numbers $s_i$ in terms of the numbers $N(a_k)$:
$$s_i=\sum\left\{N(S):S\subseteq\{a_1,\dots,a_r\}\text{ and }|S|=i\right\}\;.$$
E.g., for $i=1$ this is just $$\sum_{k=1}^rN\big(\{a_k\}\big)\;.$$
We want to prove that for $m=0,\dots,r$,
$$\begin{align*} e_k&=s_k-\binom{k+1}1s_{k+1}+\ldots+(-1)^r\binom{r}{r-m}s_r\\ &=\sum_{m=k}^r(-1)^{m-k}\binom{m}{m-k}s_m\\ &=\sum_{m=k}^r(-1)^{m-k}\binom{m}ks_m\;. \end{align*}$$
There are several ways to do this, so the one that I’m going to use may not be the one that you’ve already seen.
Suppose that $x$ is an object with at least $m$ of the $r$ properties. Then there is a $k$ such that $m\le k\le r$ and $x$ has exactly $k$ of the properties. Let $P$ be the set of $k$ properties possessed by $x$; $P$ has $\binom{k}m$ subsets of size $m$, and $x$ is counted once for each of them in the calculation of $e_m$, so $x$ contributes a total of $\binom{k}m$ to $e_m$. It follows that the $e_k$ objects with $k$ of the properties contribute a total of $\binom{k}me_k$ to $e_m$ and hence that $$s_m=\sum_{k=m}^r\binom{k}me_k\;.\tag{1}$$
Now I’ll define two polynomials: let
$$S(x)=\sum_{k=0}^rs_kx^k\quad\text{and}\quad E(x)=\sum_{k=0}^re_kx^k\;.$$
In view of $(1)$ we have
$$\begin{align*} S(x)&=\sum_{m=0}^rs_mx^m\\ &=\sum_{m=0}^r\sum_{k=m}^r\binom{k}me_kx^m\\ &=\sum_{k=0}^r\sum_{m=0}^k\binom{k}me_kx^m&&\text{reverse the order of summation}\\ &=\sum_{k=0}^re_k\sum_{m=0}^k\binom{k}mx^m\\ &=\sum_{k=0}^re_k(x+1)^k&&\text{by the binomial theorem}\\ &=E(x+1)\;. \end{align*}$$
Of course this implies that $E(x)=S(x-1)$, and we can reverse the above calculation to find that
$$\begin{align*} E(x)&=S(x-1)\\ &=\sum_{m=0}^rs_m(x-1)^m\\ &=\sum_{m=0}^rs_m\sum_{k=0}^m(-1)^{m-k}\binom{m}kx^k&&\text{binomial theorem again}\\ &=\sum_{k=0}^r\sum_{m=k}^r(-1)^{m-k}\binom{m}ks_mx^k&&\text{reverse the order again}\;. \end{align*}$$
The coefficient of $x^k$ in $E(x)$ is therefore
$$e_k=\sum_{m=k}^r(-1)^{m-k}\binom{m}ks_m\;,$$
exactly as we wanted.