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.
What you want is not the total time a customer spends in the system, but just in the line. Do you know the formula for the queue waiting time $\mathcal{W_q}$ distribution of an M/M/3 queue? Its a pretty common formula in queue theory textbooks. Its cumbersome to write, so ill jsut link you to it, see Slide 35 of this ppt. And p.13 of this
Best Answer
Your reasoning is correct as well as your answer for 1) but not for 2), due to an incorrect value for $\mathbb P(Y=1)$. Perhaps a simpler way of solving 2) is to calculate the complementary probability: $$ \mathbb P(Y=0)=\frac{5}{6}\cdot\frac{5}{6}=\frac{25}{36},\qquad \mathbb P(Y\geq 1)=1-\mathbb P(Y=0)=\frac{11}{36}. $$ Now comparing with your method, we can see that $\mathbb P(Y=1)$ does not equal $\frac16$ as you wrote, but rather $\frac{10}{36}$. How to see it directly? Call the arrival times $T_1$ and $T_2$. Then either $T_1$ falls in the last 10 minutes and $T_2$ falls in the first 50 minutes, or vice versa. Thus $$ \mathbb P(Y=1)=\frac{1}{6}\cdot \frac56 + \frac56\cdot\frac16=\frac{10}{36}. $$