(1) No (2) Yes, there is an explicit formula. If the $D_i^\prime$ resp. $D_i$ are $\exp(\mu_1)$ resp. $\exp(\mu_2)$, and the second shot is at $t$ , $$\mathbb{P}(E^c)=e^{-m\mu_1t} {\alpha^m\,B(m,\alpha(n-m)+1) \over B(m,n-m+1)}$$ where $\alpha={\mu_2 \over \mu_1 +\mu_2}$ and $B$ denotes the Beta-Function.

Your formula is not correct. Let $m=\lfloor n/2 \rfloor +1$ and denote the delay times of the first rep. second shot by $D_i^\prime$ resp. $D_i$. The correct formula is $\mathbb{P}(E^c)={n \choose m}\mathbb{P}(E^c, B=\{1,\ldots,m\}$). Let $M_m=\max \{D_1,\ldots,D_m\}$. Compute $\mathbb{P}(E^c, B=\{1,\ldots,m})$ by integrating $\mathbb{P}(D_1^\prime>t+x_1,\ldots, D_m> t+x_m, D_1=x_1,...,D_m=x_m, M_m=s, D_{m+1}>s,\ldots, D_n>s)$ (first the $x_i$ from $0$ to $s$, then $s$ from $0$ to $\infty$).

Since $a_{d,m}(k)=k![z^k] \left(\sum_{i=0}^d {z^i \over i!}\right)^m$ the asymptotics is given given by the ``large powers'' case of the saddlepoint method, see e.g. Prop. VIII 7 and Thm. VIII 8 of Flajolet and Sedgewick, Analytic Combinatorics

In "A recurrence related to trees", Proc. of the AMS (1989), Knuth and Pittel have investigated the ``tree polynomials" $t_n(y)=n![z^n]1/(1+T(z))^y$ (where $T(z)=-W(-z)$) and in praticular given the first two terms of the asymptotic expansion for fixed $y$

I should have stated that the $X_i$ are jointly Multinomial (with $n$ and $p_1=...=p_k=1/k$), which makes clear that (1) it agrees with the model of a randomly chosen hash function and (2) that each $X_i$ is $Bin(n,1/k)$. A proof along the elegant lines of the first solution can also be given, by showing that the no. $X_c$ of pwds which hash to the same value as a randomly chosen pwd has distribution $1+Bin(n-1,1/k)$, and computing $\mathbb{E}(R)=\mathbb{E}{n+1 \over 1+X_c}$. But I found the approach above simpler.

The combinatorial reason is that ${ 1\over 1-x}$ resp. ${x \over 1-x}$ are the exp. generating functions for ordered resp. non-empty ordered sets (sequences). But you can simply extract coefficients and argue directly to see that its true.

In my view there is no need to cite this. If you think it is appropriate just cite this discussion (weblink).

By Fubini's theorem you proceed as follows: (1) in the first step consider the values $0\leq r_m < r_{m+1}$ of $R_m$ resp. $R_{m+1}$ as fixed and integrate with resp. to $L$, you get $\mathbb{P}(r_m\leq L < r_{m+1})=\exp(-\lambda r_m)- \exp(-\lambda r_{m+1})=:f(r_m,r_{m+1})$ (2) in the second step you integrate $f$ with resp. to $(R_m,R_{m+1}$

1) Yes, I am referring to Fubini's theorem (as it is formulated in the link you give. It allows (under appropriate conditions) to compute a multiple integral as an iterate integral) (2) if $L$ is independent of $(R_m,R_{m+1})$ and $0\leq R_m \leq R_{m+1}$ the first step above is valid (3) thus you only have to compute $\mathbb{E}e^{-\lambda R_m}$ resp.$\mathbb{E}e^{-\lambda R_{m+1}}$ (4) for your case (independent summands) it's easy: just use the product rule for the expectation of a product independent rvs

(1) For the first step I've only used Fubini's theorem. (2) The Laplace-transform of a (nonnegative) rv $X$ is just the function $p\mapsto \mathbb{E}e^{-pX}$. (3) You can find both e.g. in W.Feller, An Intro. to Prob. Theory 2, (1970)