$1.$ 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.
Great question :-)
Let $N$ be the event that no-one was killed, and $C$ the event that Aidan is crazy. Then you want the conditional probability
\begin{align}
\textsf{Pr}(C\mid N)
&=\frac{\textsf{Pr}(C\cap N)}{\textsf{Pr}(N)}
\\
&=\frac{\textsf{Pr}(C\cap N)}{\textsf{Pr}(C\cap N)+\textsf{Pr}(\overline{C}\cap N)}\\
&=\frac{\frac12\cdot1}{\frac12\cdot1+\frac12\cdot\frac13}\\
&=\frac34\;.
\end{align}
So the chance wasn't quite as high as you thought, but it was still the right choice.
You calculated
\begin{align}
1-\textsf{Pr}(\overline{C}\cap N)&=\textsf{Pr}((\overline C\cap\overline N)\cup(C\cap N)\cup(C\cap\overline N))\\
&=\textsf{Pr}((\overline C\cap\overline N)\cup C)\;,
\end{align}
the unconditional probability that either Aidan is Mafia and someone was killed or he's crazy (and hence no-one was killed). I don't think this corresponds to any conditional probability given your knowledge of $N$.
Regarding your question "how can it be a $50\%$ chance he's a crazy person, but actually an $80\%$ chance" (in fact $75\%$), the former is the unconditional probability for him to be a crazy person and the latter is the conditional probability, given your knowledge of $N$.
(By "unconditional", I mean conditional only on the facts that you regarded as definitely settled: That the three of you were all townspeople and Aidan was either Mafia or crazy. Of course the original unconditional probability for Aidan to be crazy was only $\frac15$.)
Best Answer
Observe that the person $k$ in line has a total advancement in the bridge distributed as $S_k+k$, where $S_k\sim \mbox{NB}(k,p)$, where $p=1/2$ is the correct tile selection probability. Now, let $n=16$ be the total number of people and $m=18$ the bridge length. Then the probability of the $k$-th person traversing is \begin{align*} \mathbb{P}(S_k+k> m)= 0.407,\, 0.593,\quad \mbox{for}\quad k=9,\,10,\:\: \mbox{respectively}, \end{align*} and thus player number $10$ is the first to have more than $50\%$ chance of traversing.
Now, define the random variables $$D=\mbox{number of dead people},\quad S=\mbox{number of survivors}.$$ Then observing that $S_{k+1}$ is the sum of $S_k$ and an independent geometric random variable, say $G$, we obtain the pmf of $D$ as \begin{align*} p_D(k)=\mathbb{P}(D=k)&=\mathbb{P}(S_k+k\le m,S_{k}+k +G>m)\\ &=\sum_{n=1}^\infty \left(F_{S_k}(m-k)-F_{S_{k}}(m-k-n)\right)p(1-p)^{n-1}. \end{align*} Finally, we get that for the above parameters, $$\mathbb{E}(S)=n-\mathbb{E}(D)=n-\sum_{k=1}^n kp_D(k)=7.$$