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.
We can write $$S = \sum_{n=1}^{N}X_i$$
where $N$ is the number of customers that arrive within one hour and $X_i$ is the total service time for customer $i$.
The law of total expectation states: $E_X[X] = E_Y[E_X[X|Y]]$ and so for this case it amounts to: $$E_S[S] = E_N[E_S[S|N]] = E_N[E_X[N\times X]] = E_N[N E_X[X]]=E[N]E[X]$$.
In other words, your calculation of $E[S]$ is correct. Now to get the variance we start with:
$$Var[S] = E[S^2]-E[S]^2 = E_N[E_S[S^2|N]]-\{E_N[E_S[S|N]]\}^2$$
$$=E[Var[S|N]+E[S|N]^2]+\{E_N[E_S[S|N]]\}^2$$
$$=E[Var[S|N]]+Var[E[S|N]] = E[N]Var[X]+E[X]^2Var[N]$$
$$=E[N]E[X^2]$$
Where the last equality follows from the equality of a Possion random variable's mean and variance.
Finally, to calculate the probability, invoke the central limit theorem with your newly calculated $E[S]$ and $Var[S]$.
Best Answer
Yes, that is correct. For a Poisson process with rate $\lambda$ per unit of time, the random number of events per $t$ units of time is a Poisson random variable $X(t)$ with rate $\lambda t$, with probability mass function $$\Pr[X(t) = x] = e^{-\lambda t} \frac{(\lambda t)^x}{x!}, \quad x \in \{0, 1, 2, \ldots\}.$$ In your case, $\lambda = 9$, your units of time is $1$-hour increments, and $t = 2$ hour increments. Thus the probability that there are $12$ arrivals during a $2$-hour increment is $\Pr[X(2) = 12]$, which is what you wrote.