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
Let's assume that you will only arrive "on the minute" (e.g. at 10:05, never at 10:05 and 15 seconds).
A) When you arrive, there are $8$ possibilities: you arrive $0$ minutes before the metro leaves (i.e. when it is leaving; in this case I'll assume you can get on the metro), you arrive $1$ minute before the metro leaves, you arrive $2$ minutes before the metro leaves, ..., you arrive $7$ minutes before the metro leaves. You will have to wait more than $5$ minutes only if you arrive $6$ minutes before the metro leaves or $7$ minutes before it leaves. Since each possibility is equally likely, and $2$ of the $8$ possibilities require waiting more than $5$ minutes, the probability is $2/8$.
(Aside: your answer of $3/5$ would be the probability of waiting at least $5$ minutes)
B) This one is more complicated. Let's assume the first bus and metro arrive at the same time (say time $0$). Then a metro arrives at times $0, 8, 16, 24, 32, 40 = 0$. A bus arrives at times $0, 10, 20, 30, 40 = 0$. In both cases I treat $40$ as "wrapping around" to $0$ because that's when the full cycle of metro-bus arrivals repeats.
If both the metro and the bus leave after less than $2$ minutes of waiting, then they must leave within less than $2$ minutes of one another (for example, if they left with a $3$ minute gap between them you'd have to wait for at least $3$ minutes). The only time this happens is at minute $0=40$ in the cycle. If you want to wait for less than $2$ minutes, you'd need to arrive at either minute $39$ or minute $40$. So there's a $2/40$ probability.
If you're interested in the probability of waiting at most $2$ minutes (i.e. you're okay if you have to wait $2$ minutes, rather than less than $2$ minutes). Now you can achieve this by arriving at minute $8$, minute $30$, minute $38$, minute $39$, or minute $40$. Since each of these $5$ arrivals is equally likely, you have a probability of $5/40$.