The number of calls to receptionist per one hour has Poisson distribution with mean $1$. Find c.d.f for time $T$ of the arrival call if it is known that at the end of the second hour there is only one call arrived.
Let $X$ be the number of calls in an hour, then $X$ has a Poisson distribution with parameter $1$, Let $Y$ be the waiting time for a call (in hours). $Y$ has an exponential distribution. But how to get to $T$ and use the given condition?
For a Poisson process $N(t)$ with rate $\lambda$, conditioned on $\{N(t)=n\}$ the arrivals $(T_1,\ldots, T_n)$ have the same joint distribution as the order statistics corresponding to $n$ independent random variables uniformly distributed on $(0,t)$ - this is a well-known result that @Lee David Chung Lin mentioned in their answer.
So conditioned on $\{N(2)=1\}$, we see here that the time $T$ of the arrival is uniformly distributed over $(0,2)$.