Exponential distribution and poisson process 
Consider a post office with two clerks. Three people, A, B, and C,
  enter simultaneously. A and B go directly to the clerks, and C waits
  until either A or B leaves before he begins service. What is the
  probability that A is still in the post office after the other two
  have left when
$i)$the service time for each clerk is exactly(nonrandom) ten minutes?
$ii)$the service times are $i$ with probability $\frac{1}{3}$,
  $i=1,2,3$?
$iii)$the service times are exponential with mean $\frac{1}{\mu}$?

Let $T_1$,$T_2$ and $T_3$ the service times for $A,B,C$
$i)$ $P(T_1>t_2+t_3|t_1=t_2=t_3=10)=0$
I did not have to do this but that's okay, since this is quite intuitive
$ii)$Let $\Omega$ the space of possible values of $t_1,t_2,t_3$, then
$$t_1>t_2+t_3\Leftrightarrow t_1=3,t_2=1,t_3=1$$
$$P(T_1>t_2+t_3)=P(T_1=3)P(T_2=1)P(T_3=1)=\frac{1}{3}\frac{1}{3}\frac{1}{3}=\frac{1}{27}$$
$iii)$ Here I'm confuse anyone can help?
EDIT: The answer of $iii)$ is $\frac{1}{4}$ but I don't know how to get it
 A: 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.
A: [Returning to part (iii) after a delay of several days due to routine medical matters.]
Intuitively, in order for A to be the last to leave she must, in effect, lose
two fair and independent coin tosses: so her probability of being the last of
the three to leave is $(1/2)(1/2) = 1/4.$ Here are the details.
First, A is in a fair contest with B to
see who finishes first. She loses to B. (The contest is fair because both clerks
have the same service rate.) 
Second, after B leaves, C takes his place. By the no memory property
(also mentioned in the Comment by @Did), A is now in
her second fair contest, now with C, to see who finishes first, which
is independent of the first contest. Being the last
to leave, she loses to C.
Note: To present the essence of this intuitive argument without being too eager
to give away the exact details, my original answer was for the probability that C is the last to leave. He must only lose the
second contest mentioned above for that to occur--with probability 1/2. (On July 17, OP objected
that this answer to a somewhat different question did not agree
with the answerbook.) 
Addendum: The answers above related to (iii) do not depend on the service rate, except to require that both clerks have the same service rate. Three independent, nonoverlapping time periods are involved, which do depend on the
common service rate. Suppose each clerk has service rate $\mu = 1/2$
so that each takes on average 2 minutes to serve a customer.
Period (1). Time until the first of A or B departs. This is the minimum of two
simultaneous exponential random variables, which is itself
exponential with an average elapsed time of 1 minute. 
Period (2). Additional time until the the second customer departs. Again this is the minimum
of two service times; exponential with average 1 minute elapsed
time.
Period (3). Additional time until the third customer departs. He or she is now alone. The
no-memory property is invoked again. So the additional elapsed
time is 2 minutes.
Thus the average time before A, B, and C all finish service
is 1 + 1 + 2 = 4 minutes. This total time in not exponential.
Assuming no one else enters the queue, on average one of the clerks
is idle half of the time our three customers are in the post office.
A: Answer for Part II
Let $T_1,T_2,T_3$ be the service times with probability density of $\frac{1}{3}$.
In the problem that you have stated, it is descrete uniform with service times 1,2 or 3.  The solution hat you have given is right but on other other hand if it is continuous uniform,  and with  a pdf of $\frac{1}{3}$, the interval must be 0-3.  Given this
write out the inequality, $T_1\ge T_2+T_3 => 0<T_3\le T_1-T_2$.  Again given this,
The required probability is $= \int_{0}^{3}\int_{0}^{T_1}\int_{0}^{T_1-T_2}\frac{1}{3}.\frac{1}{3}.\frac{1}{3}dT_3dT_2dT_1 = \frac{1}{6}$.
Part III,
A different way of solving is given in the below file
http://mitran-lab.amath.unc.edu:8081/subversion/InfoTheory/ReducedModels/notes/Fall2014/Mathematica%20Results/hw6f12solns.pdf
Good luck 
Satish
A: The problem makes no mention of independence, so I'd be careful making that assumption. Let $S_A, S_B, S_C$ be the service times of each of the customers. Since they follow an exponential distribution with mean $\frac{1}{\mu}$, then the rate is simply $\mu$. You want to know:
$$
P(S_A > S_B + S_C)
$$
Condition on one of those, say like this:
$$
P(S_A > S_B + S_C) = P(S_A > S_B + S_C | S_A > S_B) \cdot P(S_A>S_B) + P(S_A > S_B + S_C | S_A < S_B) \cdot P(S_A<S_B)
$$
Then some simplification:
\begin{align*}
P(S_A > S_B + S_C) &= P(S_A > S_C) \cdot P(S_A > S_B) + 0 \\
&= P(S_C < S_A) \cdot P(S_B < S_A) \\
&= \frac{\mu}{\mu + \mu} \cdot \frac{\mu}{\mu + \mu} \\
&= \frac{\mu}{2\mu} \cdot \frac{\mu}{2\mu} \\
&= \frac{1}{4}
\end{align*}
