I actually figured it out myself and decided to post the answer in case someone needed it.
(a) It is the mean service time: μ = (1/4 + 1/2 + 1)/3 = 7/12.
(b) This is μ/(1 − λμ) = μ/(1 − μ) = 7/5 hours.
(c) This is 1/(1 + 7/5) = 5/12.
Good question. There are two mistakes here. The first is an incorrect derivation of the rate $\lambda$. The second is because $\overline{W} \neq \sum_{n=0}^{\infty}(n+1)\pi_n/3$. See below for details.
First, I agree with your computation that:
\begin{align}
&\pi_0=1/2 \\
&\pi_n=(1/3)^n \quad \forall n \in \{1, 2, 3, ...\}\\
&\overline{L} = \sum_{n=1}^{\infty}n\pi_n = 3/4
\end{align}
1) The total arrival rate $\lambda$ is actually:
$$ \lambda = 2\pi_0 + 1(1-\pi_0) = 1 + \pi_0 = 3/2 $$
Hence, by Little's formula: $\overline{W} = \frac{\overline{L}}{\lambda} = 1/2$.
2) Your alternative derivation for $\overline{W}$ implicitly assumes that the fraction of jobs that "see" $n$ jobs in the queueing system when they arrive is equal to $\pi_n$. If arrivals were Poisson, we could claim this by PASTA (Poisson Arrivals See Time Averages). However, arrivals are not Poisson.
Here is a correct alternative derivation: Let's make the arrivals Poisson of rate $\lambda_{new}=2$ by adding "virtual arrivals" of rate 1 when the system is busy. To make the new system equivalent to the old, suppose it works this way: If an arrival occurs when the system is empty, then it is a "true" arrival and it begins its service in the queueing system. If it occurs when the system is busy, with probability 1/2 it is a "true" arrival and it joins the queue. With prob 1/2 it is a "virtual" arrival and immediately exits with total delay 0. Now, PASTA holds for this system, and $\pi_n$ is the same as before for this system. So:
$$ \overline{W}_{new} = \pi_0 (1/3) + \sum_{n=1}^{\infty} \pi_n\frac{n+1}{6} = \frac{3}{8} $$
where we have used the fact that the expected delay, given we arrive when there are $n\geq 1$ jobs, is $\frac{1}{2}0 + \frac{1}{2}\frac{n+1}{3}$.
However, $\overline{W}_{new}$ can be decomposed into a weighted sum of the delay of true arrivals and virtual arrivals. The total rate due to true arrivals is $2-(1-\pi_0)= 3/2$ and the total rate due to virtual arrivals is $1-\pi_0=1/2$. Thus:
$$ \overline{W}_{new} = \frac{3/2}{2}\overline{W}_{true} + \frac{1/2}{2}0 $$
Since we already know $\overline{W}_{new}=3/8$, this gives $\overline{W}_{true} = 1/2$ which is consistent with part 1 above.
3) Why is your way of computing average time between arrivals incorrect?
You give an expression $\pi_0/2 + (1-\pi_0)/1$. This of course assumes a PASTA-like property that is not necessarily true. But even with PASTA, how do you get this expression? If you consider a job arriving to an empty system, the next inter-arrival time is not exponentially distributed with rate $1$, nor is it exponentially distributed with rate $2$.
Best Answer
Let $\lambda=60$, then the arrival rate to station $A$ is $60p$ and the arrival rate to station $B$ is $60(1-p)$, so the utilization at station $A$ is $\rho_A=p$ and the utilization at station $B$ is $\rho_B=4(1-p)$ (so we must have $p>\frac34$ for the system to be stable). The expected size in system is \begin{align} \mathbb E[L_A] &= \sum_{n=1}^\infty n(1-\rho_A)\rho_A^n= \frac{\rho_A}{1-\rho_A} = \frac p{1-p}\\ \mathbb E[L_B] &= \sum_{n=1}^\infty n(1-\rho_B)\rho_B^n= \frac{\rho_B}{1-\rho_B} = \frac{4(1-p)}{4p-3}. \end{align} By Little's law, the mean sojourn times are \begin{align} \mathbb E[W_A] &= \frac{\mathbb E[L_A]}{\lambda_A} = \left(\frac p{1-p}\right)\frac1p = \frac1{1-p}\\ \mathbb E[W_B] &= \frac{\mathbb E[L_B]}{\lambda_B} = \left(\frac{4(1-p)}{4p-3}\right)\frac1{4(1-p)} = \frac1{4p-3}. \end{align} Over time we can reasonably expect the distribution of customers to tend so that the expected waiting time is the same at each station. Therefore $$\mathbb E[W_A]=\mathbb E[W_B]\implies \frac1{1-p}=\frac1{4p-3}\implies p=\frac45. $$