There are four parts to this, I think your question concerns the first item:
1) time spent waiting for server 1 which in turn is split into two cases depending on which clerk finished first: $\frac{1}{\mu_1}\left( \frac{\mu_2}{\mu_1 + \mu_2} \right) + \left( \frac{1}{\mu_2} + \frac{1}{\mu_1} \right) \left( \frac{\mu_1}{\mu_1 + \mu_2} \right)$
But after some algebra, this is equal to $ = \frac{1}{\mu_1} \left( \frac{\mu_1 + \mu_2}{\mu_1 + \mu_2} \right) + \frac{\mu_1}{\mu_2(\mu_1 + \mu_2)} = \frac{1}{\mu_1} + \frac{\mu_1}{\mu_2(\mu_1 + \mu_2)}$
Which means that what "they" have shown you is right after all.
I think the problem with your calculation is that you forgot to add $\frac{1}{\mu_1}$ when server 1 finishes first. Remember that this process is memoryless. So when 1 finishes first, you have waited the time you spent with clerk 1, and on top of that, you will have to wait for the person being served at 2 as it they started anew.
BTW, these should be the other parts of the problem in case anyone cares:
2) Time getting service with server 1: $\frac{1}{\mu_1}$
3) Time spent waiting for server 2: $0\left(\frac{\mu_2}{\mu_1 + \mu_2}\right) + \left( \frac{1}{\mu_2} \right) \left( \frac{\mu_1}{\mu_1 + \mu_2} \right)$
4) Time spent getting service with server 2: $\frac{1}{\mu_2}$
The solution should be the addition of the 4 parts.
So, first, here is the tree of possibilities for how events will develop:
(It's not the complete tree - that would be infinite - but it contains all the relevant branching points.)
So, for example, after customer 1 arrives to server 1, one of two things could happen next: the customer could leave, or customer 2 could arrive to server 2.
To answer the questions, we also have to know the probabilities of taking each branch in this tree. These are proportional to the service and arrival rates. So, after customer 1 arrives to server 1, that customer's service time is exponential with rate $\frac13$, and the next customer's arrival time is exponential with rate $2$. Therefore the two arrows from the top node have odds $ \frac13 : 2$, and probabilities $\frac{1/3}{1/3 + 2} = \frac17$ and $\frac{2}{1/3 + 2} = \frac67$ respectively.
Similarly, if customer 1 and customer 2 are at servers 1 and 2, respectively, then three things could happen next: customer 1 could leave (rate $\frac13$), customer 2 could leave (rate $\frac14$), or customer 3 could arrive (rate $2$). So the odds of these outcomes are $\frac13 : \frac14 : 2$ and their probabilities are $\frac{1/3}{1/3+1/4+2}$, $\frac{1/4}{1/3+1/4+2}$, and $\frac{2}{1/3+1/4+2}$, respectively.
This is enough information to answer questions (a) and (c): you just decide which nodes in the tree tell you "customer 1 finishes before customer 2" or to "customer 3 won't have to wait", and then find the probability of ending up at those nodes of the tree - by multiplying the probability of taking each branch to get there.
Let's do this for question (a) as an example.
- At the top node "1 arrives to server 1", we have already found that the probabilities of going to "1 leaves" and "2 arrives to server 2" are $\frac17$ and $\frac67$.
- At the node "2 arrives to server 2", the rates of 1 leaving, 2 leaving, and 3 arriving are $\frac13$, $\frac14$, and $2$, so the probabilities of going to "1 leaves", "2 leaves", and "3 arrives and waits" are $\frac{1/3}{1/3+1/4+2} = \frac{4}{31}$, $\frac{1/4}{1/3+1/4+2} = \frac{3}{31}$, and $\frac{2}{1/3+1/4+2} = \frac{24}{31}$, respectively.
- At the node "3 arrives and waits", the probabilities of going to "1 leaves" and "2 leaves" are $\frac{1/3}{1/3+1/4} = \frac47$ and $\frac{1/4}{1/3+1/4} = \frac37$, respectively.
Customer 1 leaves first in three cases:
- we go directly to "1 leaves" from the top node, which has probability $\frac17$.
- we go from the top node to "2 arrives to server 2" then "1 leaves", which has probability $\frac67 \cdot \frac{4}{31}$.
- we go from the top node to "2 arrives at server 2" then "3 arrives and waits" then "1 leaves", which has probability $\frac67 \cdot \frac{24}{31} \cdot \frac47$.
Together, these add up to $\frac17 + \frac67\cdot\frac{4}{31} + \frac67 \cdot \frac{24}{31} \cdot \frac47 = \frac{31}{49}$.
(Note that we could ignore customer 3 and get a simpler tree without the "3 arrives and waits" branch, since that doesn't affect the probability of customer 1 leaving first. Then the calculation is simpler: $\frac17 + \frac67 \cdot \frac47 = \frac{31}{49}$.)
For question (b), there's a further step. Customer 2 takes an average of $3$ units of time in the system at server 1, or an average of $4$ units of time at server 2. So you should weight these averages by the probabilities that customer 2 will arrive at server 1, versus server 2.
Best Answer
In your question it is implicitly assumed that the system operates in the steady state. For the steady state to exist the system's load (denote it by $\rho$) must be less than one i.e. $\rho=\lambda \mu<1$.
The system you are asking about, when operating in the steady state, follows the cycle $empty \rightarrow busy \rightarrow empty \rightarrow busy \rightarrow ...$. Thus the proportion of time that the system is empty is $$ {(mean \,\, empty \,\, time) \over (mean \,\, empty \,\, time) + (mean \,\, busy \,\, time)}= {{1 \over \lambda} \over {1 \over \lambda} + \mu}. $$
Thus the arrival rate of customers which find the system empty is $\lambda \times {{1 \over \lambda} \over {1 \over \lambda} + \mu}$.
p.s. The latter conclusion uses the PASTA property of the Poisson process. So in general it is not valid.
p.p.s. in Kendall's notation, you are asking about the $M/G/1/0$ queueing system. In any classic queueing theory book, there are many results on this queue.