If you have a Poisson process with parameter $\lambda$ ($\lambda$ is the average number of events occurring in an interval of given unit length), and if $X$ is the number of events occurring in an interval of length $t$, then $X$ has Poisson distribution with parameter $\lambda t$. That is
$$
P[X=k] =(\lambda t)^k {e^{-\lambda t}\over k!},\quad k=0,1,2,\ldots
$$
Now on to your specific problem.
You are free to take the "unit time" to be whatever you like. Let's take it to be one hour.
Then, since the average number of calls every 10 minutes is 1, the parameter for the Poisson process is $\lambda=6$ (events per hour). For the problem we will need to express the length $t$ of a time interval in hours.
For the first part:
If $X$ is the number of events occurring in the first 10 minutes, then $t=1/6$ and $\lambda t=6\cdot{1\over 6}=1$; so
$$
P[X=k] =(1)^k {e^{-1}\over k!}, \quad k=0,1,2,\ldots
$$
You need to find $P[X=0]$.
For the second part of the problem, using the independence assumption for disjoint time intervals in a Poisson process, you can find the probability that exactly one call is
made in the first 5 minutes. If $Y$ is the number of calls made in the first 5 minutes, then $t=1/12$ and $\lambda t=6\cdot{1\over12}=1/2$; so
$$
P[Y=k] =(1/2)^k {e^{-1/2}\over k!}, \quad k=0,1,2,\ldots.
$$
You need to find $P[Y=1]$.
If you want the probability that both parts of your problem hold (it's not clear to me if this is what you want), by independence, you may multiply the two probabilities found above (remember, in a Poisson process, events that happen in one time period do not influence events that happen in another, disjoint time period).
The interarrival time follows an exponential distribution with parameter equal to the Poisson parameter. The proof is as follows:
Suppose $X$ is the Poisson variable with mean $3$. Then, let $T$ be the time of first arrival.
$P(T\geq t)=P($No arrival in first $t$ hours$)=P(X(t)=0)=\dfrac{e^{-3t}(3t)^0}{0!}=e^{-3t}$ which illustrates that the interarrival time follows Exponential$(3)$.
Best Answer
The answer given by @Sasha is probably the fastest way to obtain the result, but your approach is fine as well (and the only possible one in case you don't know yet that inter-arrival times are exponentially distributed).
As you have said, $\mathbb{P}(3\leq T_4\leq5) = \mathbb P(T_4\leq5)-\mathbb P(T_4<3)$. From there, we have (with $N(t)$ the number of Poisson events that have happened before $t$): \begin{align*} \mathbb P(T_4\leq 5) &= \mathbb P(N(5)\geq 4)\\ &= 1 - \mathbb P(N(5)\leq 3)\\ &= 1 - \left(1 + \frac{5}{1} + \frac{5^2}{2} + \frac{5^3}{6}\right)e^{-5}\\ &= 1 - \frac{118}{3}e^{-5}, \end{align*} and similarly \begin{align*} \mathbb P(T_4<3) &= \mathbb P(N(3)\geq 4)\\ &= 1 - \mathbb P(N(3)\leq 3)\\ &= 1 - \left(1 + \frac{3}{1} + \frac{3^2}{2} + \frac{3^3}{6}\right)e^{-3}\\ &= 1 - 13e^{-3}. \end{align*} So this approach does yield the same result $$\mathbb{P}(3\leq T_4\leq5) = 13e^{-3} - \frac{118}{3}e^{-5}.$$
I now realize that what might be troubling you is that in one case the inequality is strict, and in the other it isn't, yet the formula for both cases is the same. This is essentially because the events have probability zero of happening at a given time. This is why the strict probability and the other are equal: \begin{align*} \mathbb P(T_4\leq3) &= \mathbb P(T_4<3) + \underbrace{\mathbb P(T_4=3)}_{=0} \end{align*}