Yes. This is because we have:
$$\Pr(A \text{ wins first}) = \sum_{n \ge 0} 0.3 \cdot 0.45^n$$
which by the geometric series can be evaluated to:
$$\sum_{n \ge 0} 0.3\cdot 0.45^n = 0.3\cdot \sum_{n \ge 0} 0.45^n = 0.3 \cdot \frac1{1-0.45} =\frac{0.3}{1-0.45}$$
and the latter expression equals $\Pr(A \text{ wins}\mid \text{someone wins})$ because "$A$ wins" and "nobody wins" are mutually exclusive.
Cardinals winning the first game complicates things, since the situation becomes less symmetrical. Let $p=0.59$.
If we can find the probability distribution function of $X$, the number of games (conditional on Cards winning the first game), then finding the mean and variance is staightforward.
Easiest is to find the (conditional) probability that $X=3$. the Cards have to win the next $2$ games. This has probability $(1-p)^2$.
Next we find the probability that $X=4$. This can happen in $2$ different ways: (i) Cards win in $4$ and (ii) Cubs win in $4$.
(i) Cards win in $4$ if they lose exactly one of Games 2 or 3, and win the rest. This has probability $(2)p(1-p)(1-p)$.
(ii) Cubs win in $4$ if they win Games 2, 3, 4. This has probability $p^3$.
For the probability that $X=4$, add the numbers obtained in (i) and (ii).
Next we find the probability that $X=5$. We could do a cases analysis. But we know the answer! It is $1-\Pr(X=3)-\Pr(X=4)$.
Remark: In the post, there is, among other things, the assertion that the probability Cubs win in $4$ games is $p^3(1-p)$. This is true for unconditional probabilities. But the Cards won the first game, that's a fact. So the probability the Cubs win in $4$ games, given that fact, is simply $p^3$.
Best Answer
To summarize this: $P((AB)^k) = \left(\frac{2}{9}\right)^k$. That's correct.
Yes, but if A wins, they've only won one game in a row, so the match isn't over. If B wins, they've won two in a row and the match is over. The other outcome that results in $k$ alternating pairs and A winning is $(BA)^k A$. This means that $$ P(X=2k+1) = \left(\frac{2}{9}\right)^k \cdot \frac{2}{3} + \left(\frac{2}{9}\right)^k \cdot \frac{1}{3} = \left(\frac{2}{9}\right)^k $$
Two problems here. One is that you've skipped the matches involving an even number of games. But also, it sounds like (*) you're summing probabilities to determine an expectation, when you should be summing weighted probabilities.
Lulu points out the sequences which terminate after an even number of games: \begin{align*} P((BA)^{k-1} AA) &= \left(\frac{2}{9}\right)^{k-1} \cdot \frac{4}{9} \\ P((BA)^{k-1} BB) &= \left(\frac{2}{9}\right)^{k-1} \cdot \frac{1}{9} \end{align*} We use the exponent $k-1$ to make sure we have $2k$ games instead of $2k+2$. So $$ P(X=2k) = \left(\frac{2}{9}\right)^{k-1} \cdot \frac{5}{9} $$ Therefore, \begin{align*} E(X) &= \sum_{k=1}^\infty (2k) P(X=2k) + \sum_{k=1}^\infty (2k+1) P(X=2k+1) \\ &= \sum_{k=1}^\infty (2k) \left(\frac{2}{9}\right)^{k-1} \cdot \frac{5}{9} + \sum_{k=1}^\infty (2k+1) \left(\frac{2}{9}\right)^k \\ \end{align*} For these, if you expand and apply the identities \begin{align*} \sum_{k=0}^\infty r^k &= \frac{1}{1-r} \\ \sum_{k=1}^\infty kr^{k-1} &= \frac{1}{(1-r)^2} \\ \end{align*} (when $|r|<1$), you will see they sum to $\frac{20}{7}$.
(*) One thing about probability, and in fact a lot of math past calculus, is that there are many different objects that have to be considered at the same time. In this question you have outcomes (individual elements of the sample space), events (subsets of the sample space), probabilities (real numbers attached to outcomes or events), a random variable (functions on the sample space), and an expectation (a real number summarizing a random variable).
In the work you showed, you have a single run-on sentence in which you join all these different objects by vague verbs: (an event) “makes” (a probability) “which is” (a sum of probabilities), etc. You may begin to understand the subject better if you try to write your thoughts in simple sentences, keeping in mind what the various objects are and how exactly they are related.