Approximation of the total number of accessible microstates So, here is a system having two subsystems $\alpha$ and $\beta$ where the two subsystems can exchange energy between them, then the total number of accessible microstates of the whole system is given by, $$\Omega(E)=\sum_{E_{\alpha}}\Omega_{\alpha}(E_{\alpha})\Omega_{\beta}(E-E_{\alpha})$$
which approximation did we use to get,$$\Omega(E) \approx \Omega_{\alpha}(\tilde E_{\alpha})\Omega_{\beta}(E-\tilde E_{\alpha})$$
where, $\tilde E_{\alpha}$ is the most probable value of $E_{\alpha}$
 A: The approximation is
$$
\Omega_\alpha(\tilde{E}_\alpha) \gg \sum_{E_\alpha \ne \tilde{E}_\alpha} \Omega(E_\alpha)
$$
or in words: the number of microstates of the most occupied macrostate (which is also very close to the one having the mean energy) dominates not just some of the other macrostates, but all of them together. It is surprising at first, but when you look into it, it is indeed the case owing to the very large numbers involved. 
A: I'll put the conclusions first. Take ideal gas as an example, if you define the function $f(E_{\alpha}) = \Omega_{\alpha}(E_{\alpha}) \Omega_{\beta}(E-E_{\alpha})$. This function f(E) will look like:
$$
f(E)=E^{N}e^{-E}
$$
where N is a very large number (same order as the number of particles, around $10^{23}$), $E$ here is dimensionless and 1 unit of E correspond to $k_B T$.
The most probable state correspond to the maximum of $f(E)$. You can set $f'(E_{max})=0$ and get $E_{max}=N$. 
The task comes down to comparing $f(E_{max})$ and $\int_0^{\infty} f(E) dE$. Your summation is actually an integral because $E$ is continuous. So the approximation claims that
$$N^N e^{-N} \approx \int_0^{\infty} E^{N}e^{-E} dE = N!$$
Take log on both sides and you have
$$
ln(N!) \approx Nln(N)- N
$$
which is the famous Stirling's Approximation taught in every statistical mechanics course. If you don't believe it plug in some numbers. I tried $N=100$ and found that they differ by less than 1%.
Now to explain why $f(E)$ would take the form $E^{N}e^{-E}$. Boltzmann distribution essentially tells us $\Omega_{\beta}(E-E_{\alpha}) \propto e^{-E_{\alpha}/k_B T} $, which explains the second term. The first term comes from
$$
E = N (1/2 m v^2) \\
\Omega_{\alpha}(E) \propto (4\pi v^2)^N \propto E^N
$$
where $v$ is the average velocity of particles.
I realize it should really be $\Omega_{\alpha}(E) \propto E^{3/2 N}$ because I really should be using the degree of freedom but it doesn't change the order of $N$.
A: The approximation is not always valid, but only for systems which are big (i.e. thermodynamic limit, particle number goes to infinity).
In the thermodynamic limit one assumes that $\Omega(E)$ scales like $\Omega(E) \sim f(E/N)^N$ where N is the number of particles (compare for example to the ideal gas). Now for the thermodynamic limit fix $\varepsilon = E/N$, and therefore we get $\Omega(\varepsilon N) \sim f(\varepsilon)^N$.
Now consider two systems. We have to find a sensible way to let $N_1, N_2$ go to infinity and for simplicity I will assume that they are asymptotically equal $N_1 \sim N_2$. Therefore we get $$\Omega(\varepsilon N) = \int dE_1\; \Omega_1(E_1)\Omega_2(\varepsilon N - E_1)$$
$$ = \int d \varepsilon_1\; N f_1(\varepsilon_1)^Nf_2(\varepsilon - \varepsilon_2)^N$$
$$= N \int d \varepsilon_1\; (f_1(\varepsilon_1)f_2(\varepsilon - \varepsilon_2))^N$$.
Here comes the important part: As $N \to \infty$ the function $g(\varepsilon_1) =f_1(\varepsilon_1) f_2(\varepsilon - \varepsilon_1)$ will become more and more thinner (to get an intuition for this plot $(4x(1-x))^N$ for several $N$). Mathematically this is justified by the method of steepest descent https://en.wikipedia.org/wiki/Method_of_steepest_descent (equation (8)).
This tells us that the limit of the above integral is (up to prefactors) given by $(f_1(\bar{\varepsilon_1})f_2(\varepsilon - \bar{\varepsilon_1}))^N/N$, where $\bar{\varepsilon_1}$ is the maximum value of $g(\varepsilon_1)$.
Putting everything together we see that $\Omega(\varepsilon N) \sim \Omega_1(\bar{\varepsilon_1} N)\Omega_2((\varepsilon - \bar{\varepsilon_1}) N)$ or stating in terms of the total energies:
$$\Omega(E) \sim \Omega_1(\bar{E}_1) \Omega_2(E-\bar{E}_1).$$
A few comments:


*

*We omitted some prefactors. These are usually not important since we typically use the entropy $S = \log \Omega$ s.t. prefactors become only additive constants.

*The most important step is the one where we say that the probability distribution is basically a sharp peak. This is extremely important not only here but also for the general justification of statistical mechanics. It says that the probability that our system does not behave according to thermodynamics is very very very ... small. One could also turn the argument around: Since we do not observe any deviation from thermodynamics in our everyday life, the approximation must be valid.
A: When you are summing some numbers which are exponential in some term, this approximation applies. In our case, we consider the following sum
$$S = \sum_i \mathcal{E}_i = \sum_{E_\alpha} \Omega_\alpha(E_\alpha)\Omega_\beta(E-E_\alpha)$$
Note that when enumerating the microstates of large systems, this is going to scale exponentially, $\mathcal{E}_i \approx \mathcal{O} (\exp(N\phi_i))$. Because of this exponential dependence, we have
$$\mathcal{E}_{max} \leq S \leq N\mathcal{E}_{max}$$
Taking the log,
$$\frac{\ln{\mathcal{E}_{\max}}}{N} \leq \frac{\ln{S}}{N} \leq \frac{\ln{\mathcal{E}_{\max}}}{N} + \frac{\ln{N}}{N}$$
In the thermodynamic limit $N \to \infty$, and so $\ln{N}/N \to 0$. So our range for $S$ goes to $0$, and $S \approx \mathcal{E}_{max}$. 
Intuitively, this makes sense because if each $\phi_i$ is every very slightly smaller than $\phi_{max}$, then it must be that $e^{N\phi_i} \ll e^{N\phi_{max}}$, it is exponentially smaller. Adding a number and exponentially smaller numbers is approximately still the same number. $10^{N*100}$ and $10^{N*99}$ are different by a factor of $10^N$. 
