$1.$ 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.
Let $E$ be the probability that $A$ wins the game finishes in an even number of games, and $O$ the probability that $A$ wins the game in a odd number of games.
We have $\mathbb{P}(A \text{ wins})=E+O.$
Let us have a look at $E$. Denots $a$ a victory for $A$ and $b$ a victory for $B$ on each round.
If the game finishes in $2$ games, then it is $aa$.
If the game finishes in $4$ games, then it is $abaa$.
If the game finishes in $6$ games, then it is $ababaa$.
Now we begin to see a pattern. The probability of winning in $2k$ matches, is $(2/3)^{k+1}(1/3)^{k-1}$
This gives
$$
E=\sum_{k=1}^{\infty} (2/3)^{k+1}(1/3)^{k-1}=\frac{4}{7}
$$
I leave the odd part to you, but you should find $O=\frac{4}{21}$ for a total of $\frac{16}{21}$
Best Answer
Let $p$ be the probability of winning a round against $A$, and $q$ that against $B$, for some $p>q$.
The probabilility of winning two games in a row out of three, if you start playing $A$ is: $$P_A~{=\mathsf P(wwl\cup www\cup lww\mid ABA)\\= pq+(1-p)qp\\= (2-p)pq}$$
The probabilility of winning two games in a row out of three, if you start playing $B$ is: $$P_B~{=\mathsf P(wwl\cup www\cup lww\mid BAB)\\= qp+(1-q)pq\\= (2-q)pq}$$
Since $p>q$ , therefore $P_A<P_B$
Intuitively, you chances of winning two rounds in a row is improved if the outside rounds are the easiest to win; since you must certainly win the middle round if you wish to win the game. If you can win that, then you have two oportnities win against the other player; so make those the easiest.
Or in simpler terms: to win the game you must win against one player twice and the other player once, and the best way to manage this is to choose the easiest target as the one to play against twice.