We need to decide between minutes and hours for our unit of time. Say it is minutes. Then the mean time between arrivals is $3$ minutes. Or else, depending on the way the exponential distribution has been introduced to you, the rate is $1/3$.
Recall that an exponential distribution with parameter $\lambda$ has mean $\frac{1}{\lambda}$. So $\frac{1}{\lambda}=3$ and therefore $\lambda=\frac{1}{3}$.
A customer has just arrived. Let $X$ be the waiting time until the next customer arrives. Then $X$ has exponential distribution with parameter $\lambda=\frac{1}{3}$. For any positive $x$,
$$\Pr(X\le x)=\int_0^x \frac{1}{3}e^{-t/3}\,dt=1-e^{-x/3}.\tag{$1$}$$
Now we can answer the questions. Interpretation is needed, since there are some ambiguities in the questions.
(a) Interpret the question as saying: "Customer Alicia has just arrived. What is the probability that there will be a customer who arrives later than Alicia, but no more than $3$ minutes later." Then we want $\Pr(X\le 3)$. By $(1)$, this is $1-e^{-3/3}$.
(b) Interpret the question as saying that Alicia has just arrived, and we want the probability that there will be a gap of at least $6$ minutes until the next customer arrives. Then we want $\Pr(X\gt 6$. By $(1)$, this is $1-\Pr(X\le 6)$, which is $1-(1-e^{-6/3})$.
Remark: The exponential distribution is at best a crude model of the situation. For one thing, banks do close. For another, there is always a long line when one is in a hurry.
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
What you have is a time homogenous Poisson process with intensity $\lambda=1/10$, minutes scale. It says that the number of event in the interval $[t,t+T]$ is distributed as a Poisson RV with intensity $T\lambda$. So in your case, you find
$$ P(N\geq2)=1-P(N=0)-P(N=1)=1-\frac{e^{-\frac{30}{10}}(30/10)^0}{0!}-\frac{e^{-\frac{30}{10}}(30/10)^1}{1!}=1-4e^{-3} $$