I'll do the first question . . .
For positive integers $n,k$ with $n > 1$, let $f(n,k)$ be the probability that a $k$-step random walk on $K_n$ ends at the starting vertex.
Then we have the recursion
$$
f(n,k)
=
\begin{cases}
\;0&\text{if}\;\,k=1\\[4pt]
{\Large{\frac{1-f(n,k-1)}{n-1}}}&\text{if}\;\,k>1\\
\end{cases}
$$
Explanation:
For $k > 1$, to be at the starting vertex after $k$ steps, one needs be at a vertex other than the starting vertex after $k-1$ steps, probability $1-f(n,k-1)$, followed by a move to the starting vertex, probability ${\large{\frac{1}{n-1}}}$.
Examining the data for small values of $n,k$ a pattern becomes evident, suggesting the closed form
$$
f(n,k)
=
\frac
{1-\left(-{\Large{\frac{1}{n-1}}}\right)^{\large{k-1}}}
{n}
$$
which can then be proved by a straightforward induction on $k$.
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,
$$\mathbb{E}(S)=n-\mathbb{E}(D)=n-\sum_{k=1}^n kp_D(k)=7.$$
Best Answer
We solve the problem on the understanding that we remember bad bridges. It is not clear whether one is also expected to solve the amnesiac version.
For $i=1$ to $9$, let $X_i=1$ if we take the wrong bridge in trying to get from island $i$ to island $i+1$. Let $X_i=0$ otherwise.
The "extra" cost of a mistake in trying to get from $i$ to $i+1$ is $iX_i$. This is because if $X_i=1$ then we have spent one bridge crossing on the bad bridge, and there are $i-1$ additional bridges we must cross.
The total number of bridges is therefore $9+\sum_{i=1}^9 iX_i$.
By the linearity of expectation, the expectation of this is $9+\sum_{i=1}^9 iE(X_i)$.
Each $X_i$ has expectation $\frac{1}{2}$. Thus our expectation is $9+\frac{1}{2}\frac{(9)(10)}{2}$.