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
Part I: Intuitive answer.
This was already describe by @BruceET. I'll just say it again with some different words...
The answer is $1/3$.
The exponential distribution is "memoryless". From the time you begin being serviced, think of the two others exponential clocks as being "reset": the additional time required for service by the each of the other two is still $\mbox{Exp}(\lambda)$, and all three times are independent. Therefore you'll as equally likely to finish last as any of the two others, and since the probability that any two finish at the same time is zero, the probability you finish last is $1/3$. This generalizes to any number of tellers.
Part II. Proof.
Here's a detailed analysis, which could be made shorter (and closer in spirit to Part I), but I decided to expand on the joint distributions because it is very informative and useful for other problems.
$X_1,X_2,X_3$ are times until tellers finish with current customers.
You'll begin at time $Y_1=\min(X_1,X_2,X_3)$. Let $Y_3=\max (X_1,X_2,X_3)$, and let $Y_2$ be the remaining variable (since all three are distinct with probability $1$, $Y_2$ is defined except on an event of probability $0$).
Your service time is $T\sim \mbox{Exp}(\lambda)$, independently of all the rest. You'll be last if and only if $Y_1+T > Y_3$. Denote this event by $A$.
Consider the joint distribution of $X_1,X_2,X_3$ conditioned on $Y_1=X_1<X_2<X_3=Y_3$. The latter event has probability $1/6$, and therefore the joint density is
$$ 6 \lambda^3 e^{-\lambda x_1} e^{-\lambda x_2} e^{-\lambda x_3} {\bf 1}_{\{x_1 < x_2\}}{\bf 1}_{\{x_2<x_3}\}.$$
The can be written as
$$ 6 \lambda^3 e^{-3\lambda x_1} e^{-2\lambda (x_2-x_1) } e^{-\lambda (x_3-x_2)}{\bf 1}_{\{x_2-x_1 >0\}} {\bf 1}_{\{x_3-x_2>0\}}.$$
By considering all $6$ permutations, we conclude from the formula above that
$Y_1 \sim \mbox{Exp}(3\lambda)$;
$Y_2-Y_1\sim \mbox{Exp}(2\lambda)$; and
$Y_3-Y_2\sim \mbox{Exp}(\lambda)$; with
$Y_1,Y_2-Y_1,Y_3-Y_2$ independent.
Therefore, the event $A$, $Y_1 + T > Y_3$, can be rewritten as $Y_1 +T > Y_1 + (Y_2-Y_1) + (Y_3-Y_2)$ or $T> (Y_2 -Y_1) + (Y_3-Y_2)$.
Since $T\sim \mbox{Exp}(\lambda)$ and is independent of $Y_1,Y_2,Y_3$, we conclude that
$$P(A) = E[e^{-\lambda [(Y_2-Y_1) + (Y_3-Y_2)]}]=E[e^{-\lambda (Y_2-Y_1)}] E[e^{-\lambda(Y_3-Y_2)}].$$
Recall that the MGF of $\mbox{Exp}(\rho)$ at $t$ is equal to $ \frac{\rho}{\rho -t}$, and therefore the righthand side is
$$ \frac{2\lambda}{2\lambda +\lambda}\frac{\lambda}{\lambda + \lambda}=\frac 13 $$