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,
$$
{\rm E[\max \{ X_1 ,X_2 \} ]} = \int_0^\infty {[e^{ - \lambda _1 x} + e^{ - \lambda _2 x} - e^{ - (\lambda _1 + \lambda _2 )x} ]\,dx} = \frac{1}{{\lambda _1 }} + \frac{1}{{\lambda _2 }} - \frac{1}{{\lambda _1 + \lambda _2 }}.
$$
Given that there are $k$ customers ahead of him, the average waiting time is $k/\lambda$. This is because if $W_1,\dots, W_k$ are any random variables, then $E(W_1+\cdots+W_k)=E(W_1)+\cdots+E(W_k)$. In our case, the $k$ random variables have exponential distribution with parameter $\lambda$.
Thus the random variable $E(T|X)$ is the constant $\dfrac{1}{\lambda}$ times a Poisson random variable with parameter $a$. It follows that $E(E(T|X))=\dfrac{a}{\lambda}$.
Best Answer
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.