$1.$ 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.
There are four states to consider, namely $AC$, $CB$, $BA$, and the ending state $E$. Here $XY$ encodes that the next match is between $X$ and $Y$, whereby $X$ has scored immediately before. At the start we are in state $AC$. Denote the probability that $Z$ will win when the game is in state $XY$ by $p_Z(XY)$.
From a figure one easily reads off that one has
$$p_A(AC)={1\over2} +{1\over2}p_A(CB),\quad p_A(CB)={1\over2}p_A(BA),\quad p_A(BA)={1\over2}p_A(AC)\ .$$
Solving this system gives $$p_A(AC)={4\over7},\quad p_A(BA)={2\over7},\quad p_A(CB)={1\over7}\ .$$
Using the inherent circular symmetry of the problem we then obtain $$p_B(AC)=p_A(CB)={1\over7} ,\quad p_C(AC)=p_A(BA)={2\over7}\ .$$
Best Answer
If $A$ is number $1$ on the line (probability on that is $\frac14$) then he must win $3$ consecutive games (probability on that $p^3$).
If $A$ is number $2$ on the line (probability on that is $\frac14$) then he must win $3$ consecutive games (probability on that $p^3$).
If $A$ is number $3$ on the line (probability on that is $\frac14$) then he must win $2$ consecutive games (probability on that $p^2$).
If $A$ is number $4$ on the line (probability on that is $\frac14$) then he must win $1$ game (probability on that $p$).
Can you take it from here?