As mentioned in the comments, the answer depends very much on the model used to describe the passage times of the buses. The deterministic situation where the passage times of buses of type $k$ are $s_k+m_k\mathbb N$ for some initial passage time $s_k$ in $(0,m_k)$ is too unwieldy to be dealt with in full generality hence we now study two types of assumptions.
(1) Fully random passage times
Here the passage times of buses of type $k$ are a Poisson process of intensity $1/m_k$ and the passage times of buses of different types are independent. Then, starting at time $t_0$, the next bus of type $k$ arrives after a random time exponential with mean $m_k$ hence the waiting time $T$ is such that
$$
\mathbb P(T\gt t)=\prod_k\mathbb P(\text{no bus of type}\ k\ \text{in}\ (t_0,t_0+t))=\prod_k\mathrm e^{-t/m_k}=\mathrm e^{-t/m},
$$
where
$$
\frac1m=\sum_k\frac1{m_k}.
$$
In particular, $T$ is exponentially distributed with parameter $1/m$, hence
$$
\mathbb E(T)=m.
$$
The case $m_1=m_2=\cdots=m_n$ yields
$$
\mathbb E(T)=\frac{m_1}{n}.
$$
(2) Fully periodic passage times with random uniform initializations
Here, buses of type $k$ pass at times in $S_k+m_k\mathbb N$ where $S_k$ is uniform on $(0,m_k)$ and the random variables $(S_k)$ are independent. Now, starting at time $t_0$, the next bus of type $k$ arrives after time $t_0+t$ if $t\leqslant m_k$ and if $S_k$ is not in a subinterval of $(0,m_k)$ of lenth $t/m_k$. Thus,
$$
\mathbb P(T\gt t)=\prod_k\left(1-\frac{t}{m_k}\right),\qquad t\leqslant \bar m=\min\limits_km_k.
$$
A consequence is that
$$
\mathbb E(T)=\int_0^{+\infty}\mathbb P(T\gt t)\,\mathrm dt=\int_0^{\bar m}\prod_k\left(1-\frac{t}{m_k}\right)\,\mathrm dt.
$$
Expanding the product yields
$$
\mathbb E(T)=\sum_{i\geqslant0}(-1)^i\bar m^{i+1}\frac1{i+1}\sum_{|K|=i}\frac1{m_K},
$$
where, for every subset $K$,
$$
m_K=\prod_{k\in K}m_k.
$$
For example, time intervals $m_1$, $m_2$, $m_3$ with minimum $m_1$ yield
$$
\mathbb E(T)=m_1-\frac{m_1^2}2\left(\frac1{m_1}+\frac1{m_2}+\frac1{m_3}\right)+\frac{m_1^3}{3}\left(\frac1{m_1m_2}+\frac1{m_2m_3}+\frac1{m_3m_1}\right)-\frac{m_1^4}{4m_1m_2m_3},
$$
which can be simplified a little bit (but not much) into
$$
\mathbb E(T)=\frac{m_1}2-\frac{m_1^2}{6m_2}-\frac{m_1^2}{6m_3}+\frac{m_1^3}{12m_2m_3}.
$$
The case $m_1=m_2=\cdots=m_n$ yields
$$
\mathbb E(T)=\frac{m_1}{n+1}.
$$
Let $t$ be the instant that bus A arrives. Then Bus B must arrive in the interval $t-4, t+5$ if you want both bus at the same time. Bus B arrival is uniformly distributed on on interval of amplitude 64 minutes, hence you have to choose 9 minutes out out 64, which is probability $9/64$.
Best Answer
There is an excellent answer to the more general question, "Expected Waiting time if the are many buses, which each stop every $m_k$ minutes", can be found here.
As mentioned in the comments on this question and the linked one, the answer depends very much on the model used to describe the passage times of the buses. Here are two possible solutions, each based on slightly different assumptions, for two bus scenario, but which give quite different results.
Case 1. Fully random times.
Here the passage times of buses of type $k$ are a Poisson process of intensity $1/m_k$ and the passage times of buses of different types are independent.
Expected Time is $$\mathbb E(T) = m, \quad \textrm{where } \; \frac{1}{m} = \frac{1}{m_1}+\frac{1}{m_2}$$.
That is, $\frac{1}{m} = (1/8+1/12) =\frac{5}{24}$. So $m$ =4 minutes 48 seconds.
Case 2. Fully periodic passage times with random uniform initializations
Here, buses of type $k$ pass at times in $S_k+mkN$ where $S_k$ is uniform on $(0,m_k)$ and the random variables $(S_k)$ are independent.
$$ \mathbb E(T)=m_1-\frac{m_1^2}2\left(\frac1{m_1}+\frac1{m_2}\right)+\frac{m_1^3}{3}\left(\frac1{m_1m_2}\right), $$ thus, with $m_1 =8, m_2 = 12$, we get $$m=8-\frac{64}{2}\left(\frac{1}{8}+\frac{1}{12}\right)+\frac{512}{3}\left(\frac{1}{96}\right) = 28/9$$ So $m=$ 3 minutes 7 seconds.
Thus, we can see from the quite different results that it depends a lot on what model/assumptions you use.