Let the random variables $X$ and $Y$ denote the respective waiting times until the first customer. These two random variables have exponential distribution with parameters say $\lambda$ and $\mu$. We want the probability that $X\lt Y$.
By independence, the joint density is the product of the individual densities, so
$$\Pr(X\lt Y)=\int_{y=0}^\infty \mu e^{-\mu y}\left(\int_{x=0}^y \lambda e^{-\lambda x}\,dx\right)\,dy.$$
The integrations are not difficult.
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
One conditions on the event that exactly $1$ customer arrives in $(2,4)$. The arrival date of this unique customer is uniformly distributed on $(2,4)$ hence the number $N_{2,3}$ of customers arriving in $(2,3)$ is Bernoulli, that is, either $0$ or $1$ with equal probability. On the other hand, the number $N_{1,2}$ of customers arriving in $(1,2)$ is Poisson $\lambda$ and independent of the number of customers arriving in $(2,4)$. Finally, the number of customers arriving in $(1,3)$ is $N_{1,3}=N_{1,2}+N_{2,3}$. Thus, considering a Poisson random variable $X$ with parameter $\lambda$, $$ P(N_{1,3}=n\mid N_{2,4}=1)=\frac12P(X=n)+\frac12P(X=n-1), $$ that is, $$ P(N_{1,3}=0\mid N_{2,4}=1)=\frac12\mathrm e^{-\lambda}, $$ and, for every $n\geqslant1$, $$ P(N_{1,3}=n\mid N_{2,4}=1)=\frac12\mathrm e^{-\lambda}\left(\frac{\lambda^n}{n!}+\frac{\lambda^{n-1}}{(n-1)!}\right). $$