What you want to calculate is $P(T_2+T_3<T_1)$, where $T_i$'s are independent exponentially distributed random variables.
First, let's develop some formulas for continuous independent random variables:
$$ P(X < Y) = \int_{x<y} f_{X,Y}(x,y)\ dxdy = \int_{-\infty}^\infty\int_{-\infty}^y f_X(x)f_Y(y)\ dxdy = \int_{-\infty}^\infty F_X(y)f_Y(y)\ dy $$
$$F_{X+Y}(z) = \int_{x+y<z} f_{X,Y}(x,y)\ dxdy = \int_{-\infty}^\infty \int_{-\infty}^{z-y} f_X(x)f_Y(y)\ dxdy = \int_{-\infty}^\infty F_X(z-y)f_Y(y)\ dy$$
Using these, we can calculate $$F_{T_2+T_3}(t) = (1 - e^{-\mu t}(\mu t + 1))1_{[0,\infty\rangle}(t)$$ and finally $$P(T_2+T_3<T_1) = \frac 1 4.$$
Let me know if you need any details.
EDIT: There is a simpler way to do this, closely related to memorylessness of exponential distribution. Let $X$ be exponentially distributed, $Y$ independent of $X$ and $f_Y(y) = 0,\ y < 0$. Then $$\begin{align*} P(X>Y) &= 1 - P(X<Y) = 1 - \int_{-\infty}^\infty F_X(y)f_Y(y)\ dy \\ &= 1 - \int_{0}^\infty(1 - e^{-\mu y})f_Y(y)\ dy = \int_{-\infty}^\infty e^{-\mu y}f_Y(y)\ dy = {E}(e^{-\mu Y})\end{align*}$$ Now, assuming $X$ is exponentially distributed, and $X,Y,Z$ pairwise independent, we have $$\begin{align*} P(X> Y + Z) = {E}(e^{-\mu(Y+Z)}) = {E}(e^{-\mu Y}e^{-\mu Z}) = {E}(e^{-\mu Y}){E}(e^{-\mu Z}) = P(X>Y)(X>Z) \end{align*}$$ All we need to notice that in your exercise symmetry implies $P(X<Y) = P(X>Y) = \frac 1 2$ (and same for $Z$).
Yet another way is to directly write down integral:
$$P(X>Y+Z) = \int_{x>y+z} f_{X,Y,Z}(x,y,z)\ dxdydz = \int_{-\infty}^\infty\int_{-\infty}^\infty\int_{y+z}^\infty f_X(x)f_Y(y)f_Z(z)\ dxdydz \\= \int_{-\infty}^\infty\int_{-\infty}^\infty (1-F_X(y+z))f_Y(y)f_Z(z)\ dydz$$ and calculate.
The mistake is that the waiting time is not distributed in the way that you have proposed, because the waiting time is the lesser of the service times of $B$ and $C$. This is because $A$ takes the first available opening, so whoever finishes first, $A$ chooses that clerk. That is to say, if $W_A$ is the waiting time of $A$ and $S_B$, $S_C$ are the service times of $B$ and $C$ respectively, then $$W_A = \min(S_B, S_C),$$ and it is not generally true that $\operatorname{E}[W_A] = \operatorname{E}[S_B]/2$.
Rather, you must compute the minimum order statistic of $B$ and $C$'s service times. Since these are iid exponentially distributed with rate $\lambda$, it follows that $$\Pr[W_A > t] = \Pr[\min(S_B, S_C) > t] = \Pr[(S_B > t) \cap (S_C > t)] \overset{\text{ind}}{=} \Pr[S_B > t] \Pr[S_C > t] \overset{\text{id}}{=} (e^{-\lambda t})^2.$$ That is to say, the probability that $A$ waits more than $t$ to begin service is the product of the probabilities that both $B$ and $C$ take more than $t$ to be finished. And this makes perfect intuitive sense--because if $A$ must wait more than $t$, then that means both $B$ and $C$ are still occupied, which means their service times are also more than $t$.
Now that you know $\Pr[W_A > t] = e^{-2\lambda t}$, what can you say about $\operatorname{E}[W_A]$?
Best Answer
After a while one of $B$ or $C$ is through, say $C$. Now $A$ goes up to the free clerk.
By the memorylessness of the exponential, the additional waiting time for $B$, no matter how long she has already spent being served, has the same exponential distribution as if $B$ had just started being served. So the probability that $B$ finishes before $A$ is $1/2$.
Remark: This is somewhat counterintuitive. And rightly so, because real waiting times do not have exponential distribution. But often the exponential model fits reasonably well.
Because these things can be difficult to see, we look at a discrete analogue of the problem. The discrete analogue of an exponentially distributed random variable is a random variabble with geometric distributions.
There are two $20$-sided dice, with faces numbered $1$ to $20$. Customers $B$ and $C$ each are flipping their die, once a minute. Whenever one of them gets a $1$, she can go home.
Now $A$ walks in. When a die becomes free, she has to toss it until she gets a $1$. What is the probability $A$ is last to go home?
This is a bit more complicated, because the two dice could become free at the same time. But that's fairly unlikely. If only one die becomes free first, say the one used by $C$, then $A$ and $B$ are "equal." The time $B$ has spent fruitlessly tossing doesn't help her go home before $A$. So the probability $A$ is last to go home is about $1/2$. A little more, because of the possibility $B$ and $C$ got their $1$'s simultaneously.
For (b), we need to know the parameter $\lambda$ of the exponential. The expected time $A$ spends is the sum of the expected time $A$ waits before a clerk is free, plus the expected time for $A$ to be processed once a clerk is free. The latter is $\frac{1}{\lambda}$. Now you can look for an argument, intuitive or more formal, about the expected time $A$ waits. For the more formal, note that the probability that neither $B$ nor $C$ is finished after time $t$ is $(e^{-\lambda t})(e^{-\lambda t})$.