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 $k=1000/e\approx \boxed{368}$.