The first scenario can be answered on base of symmetry: $\frac{1}{2}$ is okay.
Let $p$ denote the probability of winning a game and $q=1-p$ the
probability of loosing it.
Probability of winning the match in $3$ games: $p^{3}$
Probability of winning the match in $4$ games: $\binom{3}{2}p^{3}q=3p^{3}q$
Probability of winning the match in $5$ games: $\binom{4}{2}p^{3}q^{2}=6p^{3}q^{2}$
So the total probability of winning is: $$p^{3}\left[1+3q+6q^{2}\right]$$
Check it for $p=0.6$ (hence $q=0.4$).
Two provide a little bit of insight, assume for simplicity that the team who wins two (not three) games wins the series.
Simplification
Since you have three games with two possible outcomes each, it is enough to consider a probability with $2^3=8$ elements. Namely
$$
\Omega = \{\omega_{AAA}, \ \omega_{AAB}, \ \omega_{ABA}, \ \dots, \ \omega_{BBB}\}
$$
For example, $\omega_{ABB}$ denotes the event that $A$ wins the first game and $B$ wins the second and third. Now define a random variable $X$ like this:
$$
X=\begin{cases}1, & \text{A wins the series}, \\0, & \text{B wins the series} \end{cases}
$$
Using the notation from above, we see that
$$
X(\omega) =\begin{cases} 1,& \omega \in \{\omega_{AAB}, \ \omega_{ABA},\ \omega_{BAA}, \ \omega_{AAA} \}=:\Omega_A \\ 0, & \omega \in \{\omega_{BBA},\ \omega_{BAB},\ \omega_{ABB}, \ \omega_{BBB} \}:=\Omega_B\end{cases}
$$
Finally we can calculate the desired probability, therefore we define the set of all the $\omega$, where $A$ wins the first game:
$$
C:=\{\omega_{AAA}, \ \omega_{AAB},\ \omega_{ABA}, \ \omega_{ABB} \}.
$$
And now:
$$
\mathbb{P}(\text{A wins series} \:| \text{A wins first game})= \mathbb{P}(X=1\: |\: C) = \mathbb{P}(\Omega_A\: |\: C) =\frac{\mathbb{P}(C\cap \Omega_A)}{\mathbb{P}(C)} = \\
\frac{\mathbb{P}(\{\omega_{AAB}, \ \omega_{ABA},\ \omega_{AAA}\ \})}{\mathbb{P}(C)},
$$
which can be calculated easily using the additivity of $\mathbb{P}$. For example
$$
\mathbb{P}(C) =\mathbb{P}(\omega_{AAA})+\mathbb{P}(\omega_{AAB})+\mathbb{P}(\omega_{ABA})+\mathbb{P}(\omega_{ABB})=\\
p^3+p^2(1-p)+p^2(1-p)+p(1-p)^2.
$$
General Case
For the general case you need to consider a probability space with $2^5 =32$ elements. Then $\Omega$ looks like this:
$$
\Omega= \{\omega_{AAAAA}, \ \omega_{AAAAB}, \dots, \ \omega_{BBBBB}\}.
$$
The interpretation of the $\omega$s is the same as before: e.g., $\omega_{AABBA}$ describes the event that $A$ wins the first two games and the last one, and $B$ wins games 3 and 4.
The random variable $X$ and the set $C$ can also be defined as before: the definition of $X$ yields a partition of $\Omega$ into two set $\Omega_A$ and $\Omega_B$, where $\Omega_A$ contains all events where $A$ wins more often than $B$ and $\Omega_B$ contains all events where $B$ wins more often than $A$.
$C$ is again the set of all events where $A$ wins the first game.
$$
C=\{\omega_{AAAAA}, \ \omega_{AAAAB}, \dots, \ \omega_{ABBBB}\}
$$
And therefore
$$
\mathbb{P}(\text{A wins series} \:| \text{A wins first game})= \mathbb{P}(X=1\: |\: C) = \mathbb{P}(\Omega_A\: |\: C) =\frac{\mathbb{P}(C\cap \Omega_A)}{\mathbb{P}(C)}
$$
Final Coment
Finally, I want to say that for exercises like this, it is way faster to compute the desired probabilty directly, without going into to much detail about the probability space in the shadows. But it is a nice exercise after all.
Best Answer
$T$ is the probability that Calvin wins from a tie. Suppose Calvin and Hobbes are tied. With probability $p$ he will win the match, at which point he will have probability $A$ of winning. With probability $1-p$ he will lose the match, at which point he will have probability $A$ of winning.
$A$ is the probability that Calvin wins once he is ahead. Suppose Calvin is ahead. With probability $p$ he can win the match and thus the whole game. With probability $1-p$ he will lose the match and then once again have probability $T$ to win.
$B$ is the probability Calvin will win once he is behind. Suppose Calvin is behind. He must win the match to win the game. He has probability $p$ to win the match, at which point he has probability $T$ to win the game.
You ask why these expressions do not contain $p^2$ or $p^3$ terms. If you like, there is a $p^2$ term that pops out if you plug the expression for $A$ into the expression for $T$. However, the expression for $T$ will then also have terms containing $T$ and $A$. Repeatedly plugging expressions into each other does not lead to a simple computation of $T$. This is analogous to what would happen if you tried to consider all of the infinitely many different possible sequences of wins and losses that might occur, and calculate each of their probabilities. The method suggested by user694818 greatly simplifies this situation.