Since the person who lost their pass plays a special role, I find it easier to think about the problem in terms of $1$ loser and $n-1$ keepers.
If the loser happens to take their own seat, everyone else also gets to sit in their own seat, so the probability for any set of people including the loser to end up in their own seats is just $1/n$.
The probabilities of various keepers sitting in their own seats are independent, so the probability for any set of keepers to end up in their own seats is just the product of the individual probabilities.
To see this, let's look at the derivation of the individual probabilities. To simplify things, I'll number the keepers in reverse order of boarding, so keeper $1$ boards last, and keeper $j$ corresponds to your $k=n-j+1$.
The reason keeper $j$ has probability $j\,/\,(j+1)$ of sitting in their own seat is that at exactly one point before keeper $j$ boards, exactly one passenger will choose either the loser's seat or one of the seats of keepers $1$ through $j$, and keeper $j$ will lose their seat iff this choice falls on their seat out of these $j+1$ seats.
Now the probability of $j'\gt j$ taking $j$'s seat certainly depends on whether $j'$ gets their own seat – but that has no bearing on the derivation above. It makes no difference to $j$ whether it's $j'$ who makes that choice and potentially takes $j$'s seat, or someone else before or after $j'$. Independent of whether $j'$ gets their own seat, exactly one person will make that choice, and hence the probabilities for $j$ and $j'$ to get their own seats are independent.
This argument also carries through for more than two keepers, and thus, as claimed, the probability for any set of keepers to end up in their own seats is the product of the individual probabilities. For instance, the probability of all keepers to end up in their own seats is the product of all individual probabilities, which telescopes and yields $1/n$, the probability of the loser choosing their own seat (as it must).
Your reasoning for part $3$ is completely correct (though the equation $H_n-1=\log n-1$ is only approximately true).
Hint for part $4$:
You should enumerate all of the ways that Joe can move exactly once, then add up the probabilities of each occurring. Let us number all $n-1$ people besides Joe from $1$ to $n-1$, in the order they sit down. It could be the case that Joe initially takes person $1$'s seat, and then moves to Joe's seat. The probability of this is $\frac1{n}\cdot \frac1{n-1}$. What are the other ways, and the other probabilities?
Best Answer
Suppose that the first $n$ passengers have lost their boarding passes. Consider the $n+1$ seats consisting of the $n$ seats assigned to the first $n$ passengers, along with the last person's seat. The order in which these $n+1$ seats get filled is entirely random, as nobody will take any of these seats based on what their boarding pass says. The last passenger will get to sit in her correct seat if and only if that seat is the last of the $n+1$ seats to get filled, so the probability that the last passenger gets her correct seat is $$ \frac{1}{n+1}. $$