For part a), you were almost right. The correct solution is
$$
\frac{1}{{\lambda _1 + \lambda _2 }} + \bigg[\frac{{\lambda _1 }}{{\lambda {}_1 + \lambda _2 }}\frac{1}{{\lambda _1 }} + \frac{{\lambda _2 }}{{\lambda {}_1 + \lambda _2 }}\frac{1}{{\lambda _2 }}\bigg] = \frac{3}{{\lambda _1 + \lambda _2 }} = \frac{3}{{1/3 + 1/6}} = 6,
$$
where we have used the following facts. If $X_i$, $i=1,2$, are independent exponential$(\lambda_i)$ random variables (meaning that they have densities $\lambda_i e^{-\lambda_i x}$, $x > 0$), then $U:=\min\{X_1,X_2\}$ is exponential$(\lambda_1+\lambda_2)$ (and hence its mean is $1/(\lambda_1+\lambda_2)$, which corresponds to the first term above), and moreover, $U$ is independent of the random variable $N$ defined by
$N=1$ if $X_1 < X_2$, and $N=2$ if $X_2 \leq X_1$, for which it holds ${\rm P}(N = 1) = \lambda _1 /(\lambda _1 + \lambda _2 )$ and ${\rm P}(N = 2) = \lambda _2 /(\lambda _1 + \lambda _2 )$. For these facts, see this post (parts (a)-(c)).
EDIT:
For part b), consider
$$
2 + 2 + \frac{2}{3}6 + \frac{1}{3}3 = 9,
$$
or more generally,
$$
\frac{1}{{\lambda _1 + \lambda _2 }} + \frac{1}{{\lambda _1 + \lambda _2 }} + \frac{{\lambda _1 }}{{\lambda _1 + \lambda _2 }}\frac{1}{{\lambda _2 }} + \frac{{\lambda _2 }}{{\lambda _1 + \lambda _2 }}\frac{1}{{\lambda _1 }} = \frac{{2\lambda _1 \lambda _2 + \lambda _1^2 + \lambda _2^2 }}{{(\lambda _1 + \lambda _2 )\lambda _1 \lambda _2 }} = \frac{{\lambda _1 + \lambda _2 }}{{\lambda _1 \lambda _2 }}.
$$
(Setting $\lambda_1 = 1/3$ and $\lambda_2 = 1/6$ gives the desired answer, $9$.)
Apparently, you were supposed to solve part b) using the above method. Nevertheless, it may be worth giving here the following alternative derivation:
$$
\frac{1}{{\lambda _1 + \lambda _2 }} + {\rm E[\max \{ X_1 ,X_2 \} ]} = \frac{1}{{\lambda _1 + \lambda _2 }} + \bigg[
\frac{1}{{\lambda _1 }} + \frac{1}{{\lambda _2 }} - \frac{1}{{\lambda _1 + \lambda _2 }}\bigg] = \frac{1}{{\lambda _1 }} + \frac{1}{{\lambda _2 }} = 9.
$$
The expression for ${\rm E[\max \{ X_1 ,X_2 \} ]}$ can be derived as follows. First note that
$$
{\rm E[\max \{ X_1 ,X_2 \} ]} = \int_0^\infty {{\rm P}(\max \{ X_1 ,X_2 \} > x)\,dx} = \int_0^\infty {[1 - {\rm P}(\max \{ X_1 ,X_2 \} \le x)} ]\,dx.
$$
Now, using the independence of $X_1$ and $X_2$,
$$
{\rm P}(\max \{ X_1 ,X_2 \} \le x) = {\rm P}(X_1 \le x){\rm P}(X_2 \le x) = (1 - e^{ - \lambda _1 x} )(1 - e^{ - \lambda _2 x} ),
$$
and hence
$$
1 - {\rm P}(\max \{ X_1 ,X_2 \} \le x) = e^{ - \lambda _1 x} + e^{ - \lambda _2 x} - e^{ - (\lambda _1 + \lambda _2 )x} .
$$
Finally,
$$
{\rm E[\max \{ X_1 ,X_2 \} ]} = \int_0^\infty {[e^{ - \lambda _1 x} + e^{ - \lambda _2 x} - e^{ - (\lambda _1 + \lambda _2 )x} ]\,dx} = \frac{1}{{\lambda _1 }} + \frac{1}{{\lambda _2 }} - \frac{1}{{\lambda _1 + \lambda _2 }}.
$$
The parameter called rate is indeed the one with the usual name $\lambda$. And the mean of the distribution with density function $\lambda e^{-\lambda t}$ (for $t\ge 0$) is $\frac{1}{\lambda}$.
This makes the choice of name $\mu$ for the rate surprising, since $\mu$ is a common name for the mean.
By the memorylessness property of the exponential, the time $X_0$ until the person currently being served is finished has exponential distribution with mean $\frac{1}{\mu}$. And the times of service $X_1$, $X_2$, $X_3$, and $X_4$ of the people waiting in line have mean $\frac{1}{\mu}$. And the time $X_5$ from the instant you get to the teller to the time you are finished has the same mean.
So your expected total time at the bank is $E(X_0+X_1+\cdots +X_4+X_5)$. By the linearity of expectation, this is $\frac{6}{\mu}$.
Best Answer
For a complete answer, it would be helpful to know where the question is taken from.
As far as I would say, an "arrival" describes the time until the next person enters the bank. As you pointed out, this is independent of the number of customers in the bank and therefore the expected time until an arrival is $1/\lambda$.
For your second question, you have to consider the number of customers being served. If all three employees are working (as it would be the case if 10 costumers are inside the bank), then the time until the next customer is leaving is described by $\min(Exp(\mu),Exp(\mu),Exp(\mu))$, where all three random variables are independent. This random variable is $Exp(3\mu)$-distributed and therefore the expected time until a departure is $1/3\mu$. Observe, this depends on the number of employees working and therefore indirectly on the number of customers inside the bank.