Disclaimer: The solutions we've shared are just one exciting approach, and there are surely many other wonderful methods out there. We’d love to hear your alternative solutions in the community thread below, so let's keep the creativity flowing!

a.

As the prospects are numbered randomly, every prospect has the same probability of $\boxed{\frac{1}{1000}}$ for having the highest rank.

b.

The $i$th prospect ($1\leq i\leq m$) has a probability of $\frac 1m$ for having the highest rank among the first $m$ prospects. So, the probability that the highest ranking prospect (among the first $m$) is one of the first $k$ prospects is $\boxed{\frac km}$. 

c.

The probability that the $(m+1)$th prospect is the highest ranking is $\frac{1}{1000}$. And the probability that this prospect will be picked is $\frac km$, since he will be picked only if the highest ranking prospect among the first $m$ prospects was one of the first $k$ who were rejected. The probability that both these events occur is therefore $\boxed{\frac{k}{1000m}}$.

d.

Note that the sum is \begin{align*} \sum_{m=k}^{999} \frac{k}{1000m} & =\frac{k}{1000}\sum_{m=k}^{999}\frac 1m \\ & = \frac{k}{1000}\left(\sum_{m=1}^{999} \frac 1m-\sum_{m=1}^{k-1} \frac 1m \right) \\ & \approx \frac{k}{1000} (\ln 1000-\ln k) \\ & = \boxed{\frac{k}{1000}\ln \left (\frac{1000}k\right)} .\end{align*}

e.

We need to find the value of $k$ for which $\frac{k}{1000}\ln \left (\frac{1000}k\right)$ is maximized. Let $f(x)=\frac{\ln x}x$. We can find that $f'(x)=\frac{1-\ln x}{x^2}$ by using the quotient rule from calculus. We know from calculus that a "good" function has its maximum when its derivative is $0$ (or in the endpoints, but we don't need that). Clearly $$f'(x)=0 \iff 1-\ln x=0 \iff x=e.$$ So the maximum is attained when $x=e$. Now notice that the function we care about is $f(1000/k)$. This is maximized when $1000/k=e$ or $k=1000/e\approx \boxed{368}$.