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,
$$
{\rm E[\max \{ X_1 ,X_2 \} ]} = \int_0^\infty {[e^{ - \lambda _1 x} + e^{ - \lambda _2 x} - e^{ - (\lambda _1 + \lambda _2 )x} ]\,dx} = \frac{1}{{\lambda _1 }} + \frac{1}{{\lambda _2 }} - \frac{1}{{\lambda _1 + \lambda _2 }}.
$$
Because you imply there are people queued, I assume $n > m.$
Each person in line begins service when the next teller
becomes available. That happens at rate $m/\lambda,$
so the mean waiting time in the queue for the next customer
is $\lambda/m.$ The mean time until that customer finishes
being served is $\lambda/m + \lambda.$
Notes: (1) To avoid potential confusion, texts on queueing models typically
use $\lambda$ for the rate of arrival of new customers, and
$\mu$ for the service rate. Means of exponential distributions
are the reciprocals of rates.
(2) The distribution of the waiting time for the minimum of
$k$ exponential waiting times (next available server) is
exponential with a rate of $k$ times the rate for
individual servers (assuming all rates equal).
(3) If you want to look for more on this topic, I suggest you
read about $M/M/k$ queues. The first M stands for (Markov,
memoryless, or exponential) interarrival times, the second
M for exponential service times, and the k for the number
of servers.
Best Answer
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 $\frac{6}{\mu}$.