The superposition of independent Poisson processes is equivalent to a single Poisson process whose rate is the sum of the rates of the independent Poisson processes and where one affects independently each point to one of the Poisson processes with a probability proportional to its rate. Thus the global interarrival time is exponentially distributed with rate the sum of the rates and the distribution of the global service time is the barycenter of the distributions of the individual service times.
See here (pages 7-10) or Kingman's classical small book Poisson Processes.
We can mechanically use the defining formula for conditional probability. As an exercise, we will go through the process. But the ultimate result is so simple that it invites further thinking.
Let $B$ be the event there were $12$ in the first two hours, and let $A$ be the event there were $5$ in the first hour. We want $\Pr(A|B)$. This is $\frac{\Pr(A\cap B)}{\Pr(B)}$.
Calculate. The number of arrivals in $2$ hours is Poisson with parameter $10$, so $\Pr(B)=e^{-10}\frac{10^{12}}{12!}$.
For the event $A\cap B$, note this happens if there are $5$ in the first and $7$ in the second. This has probability $\left(e^{-5}\frac{5^{5}}{5!}\right)\left( e^{-5}\frac{5^{7}}{7!}\right)$.
Divide. The powers of $e$ stuff cancels, and we end up with something that may be very familiar to you from your experience with the binomial distribution. With some manipulation, we arrive at
$$\binom{12}{5}\left(\frac{1}{2}\right)^{12}.$$
Added: Such a simple answer deserves a simple explanation. We had $12$ occurrences of a customer arriving. Label these $12$ customers in an arbitrary random way. Call a customer prompt if she arrives in the first hour. The probability that a customer randomly chosen from the $12$ is prompt is $\frac{1}{2}$, since she is equally likely to have arrived in the first hour as in the second. So the probability that exactly $5$ of the $12$ customers are prompt is $\binom{12}{5}\left(\frac{1}{2}\right)^{12}$.
Remarks: $1.$ You should track down the general case. We have a Poisson with rate $\lambda$ (per hour). Given that there were $n$ arrivals in the first $a+b$ hours, what is the probability that there are $k$ arivals in the first $a$ hours? Go through the same conditional probability calculation. You will get a familiar "binomial-type" expression. Now explain why the result is "obvious," and the conditional probability machinery was not necessary.
$2.$ Given that there were $12$ successes in $2$ hours, the event that there were $5$ in the first hour sounds not unlikely. So the number obtained in the post is too small by several orders of magnitude.
Best Answer
Time between arrivals is exponential with mean $1/\lambda=1/3$ hour. Service rate $\mu=2$ per hour. Length of a service time is exponential with mean $1/\mu=0.5$ hours.
This is an $M/M/2/3$ queue. 2 servers with $N=3$ max capacity which means a waiting room of size $1.$
You can use the total number currently in the system as the state space (as opposed to a vector keeping track of which server is busy.) This gives:
state $= 0$ means no one in system.
state $= 1$ means 1 customer being served, no one waiting
state $= 2$ means 2 customers being served, no one waiting
state $= 3$ means 2 customers being served, 1 waiting.
no other states possible.
We want the stationary distribution. It is a birth and death process so we know the answer from the general case. Or we can use the "balance equations." (Tricky part: departure rate in a state is proportional to the number of busy servers.)
After some algebra, $\pi_0=32/143$ and $\pi_3=27/143.$ Check my work.
Now the probability that an arrival is turned away since the system is full is $\pi_3.$ So then we know the proportion of arrivals that are able to get into the system. (Technical point: this is because the PASTA property holds here.)