$1.$ This is easy, the fact you didn't answer it correctly is probably due to not understanding the question. Person $1$, let's call her Alice, wins the tournament in $3$ games if she wins Games $1$, $2$, and $3$. We are assuming independence, so the probability is $(0.6)^3$.
$2.$ The tournament lasts exactly $3$ games if Alice wins Games $1$, $2$, and $3$, or Betty does. The probability Betty does is $(0.4)^3$, so our required probability is $(0.6)^3+(0.4)^3$.
$3.$ This is more complicated. The tournament lasts exactly $4$ games if (i) Alice wins the $4$th game, and exactly $2$ of the other $3$ or (ii) Betty wins the $4$th game, and exactly $2$ of the other $3$.
For (i), winning $2$ of the first $3$ can happen in the patterns WWL, WLW, and LWW. Each of these has probability $(0.6)^2(0.4)$. Multiply by $3$ because of the $3$ different ways. We get $3(0.6)^2(0.4)$. Multiply by the probability Alice wins the $4$th game. We get $3(0.6)^3(0.4)$.
We get a similar expression for (ii), reversing the roles of $0.6$ and $0.4$. Add: we get
$$3(0.6)^3(0.4)+3(0.4)^3(0.6).$$
$4.$ This is a sum of the probabilities that Alice wins in $3$, in $4$, and in $5$. We already know the answers to the first two: $(0.6)^3$ and $3(0.6)^3(0.4)$ respectively. I will leave to you to find the probability Alice wins in $5$. Hint: She has to win the $5$th game, and exactly $2$ of the first $4$.
$5.$ Hint: It has something to do with the answers to $1$ and $4$. The key word is conditional probability.
In symbols, let $A$ be the event "Tournament lasts $3$ games" and
$B$ the event "Alice wins tournament." We want $\Pr(A|B)$.
$6.$ Again, a conditional probability.
After you have worked on the problems for a while, perhaps I can add to the hints. Would need to know what you have been exposed to about conditional probability.
Remark: In case you are not familiar with the tournament setup, here is an explanation of how it works. As soon as one person has won $3$ games, the tournament is over. So the tournament can last $3$, $4$, or $5$ games. If some person wins Games $1$, $2$, and $3$, the tournament is over, no more games are played.
You will have to sum up a number of mutually exclusive probabilities:
The last winning match has to be $B$, and $B$ must win any $2$ of the preceding ones,
hence $\binom22 + \binom32 + \binom42$ cases
$BBB,\;ABBB,\; BABB,\; BBAB,\; AABBB,\; ABABB,\; ......$
with probabilities $0.3*0.8*0.3 + 0.7*0.8*0.3*0.8 + .....$
Best Answer
You are computing the probability that "A wins at the 6th game"
This is not the same as the probability that "A wins at the 6th game, given that the series ends at the 6th game"
In fact:
$$P(A\mbox{ wins at 6th g. | series ends at 6th g.})=\frac{P(A\mbox{ wins at 6th g. and series ends at 6th g.})}{P(\mbox{series ends at 6th g.})}$$ $$=\frac{P(A\mbox{ wins at 6th g.})}{P(\mbox{series ends at 6th g.})}$$ $$=\frac{P(A\mbox{ wins at 6th g.})}{P(A\mbox{ wins at 6th})+P(B\mbox{ wins at 6th)}}$$ $$=\frac{5/32}{5/32+5/32}=1/2$$
Of course your "intuitive" argument would suffice and is rigorous enough, since exchanging "$A$" with "$B$" does not change the probabilities.