Conditional Probability and Shark Attacks

That is correct; for completeness you might want to add your derivation of the final answer (cancelling stuff from fractions). But otherwise, that's correct!

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Yup, that's correct! I was actually going to point that out as an interesting side effect of the conclusion, but by the time I noticed I'd already posted the reply and didn't want to self-comment or edit the post. But that is correct, because the sum of the probabilities that you get attacked on days $n=1,2,3,\ldots$ is $1$ so you will always get attacked at some point.

Yes.   The probability for the first attack being on day $n$ is clearly the probability for being attacked on that day and not attacked on any of the prior days, and you have calculated that well.

$$\begin{align}\mathsf P(N=n)&=\dfrac{1}{n+1}\prod\limits_{k=1}^{n-1}\dfrac{k}{k+1}\\[1ex]&=\dfrac 1{n+1}~\dfrac{n-1}{n}\dfrac{n-2}{n-1}\cdots\dfrac{1}{2}\\[1ex]&=\dfrac{1}{n~(n+1)}\end{align}$$