Answer:
A) Single-server single-phase model (M/M/1).

B) The goal is not met, as the average time waiting for service is 5.56 minutes.
C) The new mean service rate is 7.5 customers/hour.
In this case, the average time waiting for service is 4 minutes, so the goal is met.
Explanation:
A) This situation can be modeled as a single-server single-phase model (M/M/1).
The mean arrival rate is 2.5 customers per hour.

The mean service rate is 6 customers per hour, calculated as:

B) The average waiting time for a customer can be expressed as:

The average waiting time is 5.56 minutes, so it is more than the goal of 5 minutes.
C) If the average time spent per customer to 8 minutes, the mean service rate becomes

An the average waiting time for the service now becomes:

The average time is now 4 minutes, so the goal is achieved.