Thanks to the memoryless property of the exponential distribution, the analysis of your system is quite easy. It is basically a modification of the M/M/2: indeed you have two servers, working in parallel with different speed if they are not empty.
Consider the state space $S=\{0,1_a,1_b,2,3,\dots\}$, where state $s$ denotes that there are $s$ customers in the system. We need a double state $1$ because we need to distinguish the two possible situations with only one customer, that could be in box 1 (a) or 2 (b). From your assumptions we know that if the system is empty, a new customer will go directly to box 1, but when there are two customers in the systems and one of them finishes the service, we may be in the situation where there is a customer in box 2 and none in box 1.
The transition rates for your system are:
- $q_{0,1a}= \lambda$,
- $q_{1a,2}=\lambda$ and $q_{1_a,0}=\mu_1$,
- $q_{1b,2}=\lambda$ and $q_{1_b,0}=\mu_2$,
- $q_{2,3}=\lambda$, $q_{2,1_a}=\mu_2$ and $q_{2,1_b}=\mu_1$
- $q_{n,n+1}=\lambda$ and $q_{n,n-1}=\mu_1+\mu_2$ for $n\geq 3$.
Notice that we need to specify clearly all the possible transitions when there are two or less customers in the system, while when they are more than three in not necessary anymore: this is because when both servers are working, one of the two will finish first and the residual service time of the other one will be again exponential with the same mean, thanks to the memoryless property of the exponential distribution.
Once one has set the mathematical problem in this way, it should not be hard to derive the number of people in the system and the waiting time distribution.