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).
I believe you can solve using either.
Using Poisson
Let $X$ be the number of calls in $30$ seconds. Then $X \sim Poisson(\lambda = 1.1)$ and $P(X=0) = e^{-1.1}$
Using Exponential
Let $Y$ be the time until the first call. The average time to the first call is $60/2.2 = 27.27$ seconds.
That means $E[Y] = 27.27$ which means $\frac{1}{\lambda} = 27.27$ so $\lambda = .0366$ and then $P(Y > 30) = 1-P(Y \le 30) = 1-(1-e^{-30 *.0366}) = e^{-1.1}$
I think the key thing is that $\lambda$ is a rate parameter so if someone says $2.2$ every minute you can chop it up to suit your needs... I am learning this stuff as well so let me know if you have any doubts.
Best Answer
We weren't given a conditional probability question. It isn't previously known or given that $6$ calls happened in the first minute. Had it said something along those lines then your conditional probability approach would have been correct.
Instead we have two events $A$ and $B$, and we want the probability of $A\cap B$, and because events over disjoint time intervals are independent in the Poisson process, we can find $\mathbb{P}(A\cap B) = \mathbb{P}(A) \mathbb{P}(B)$