The unstable region for the terminating network attached to port 2 is essentially the left side of the Smith Chart. We cannot choose a value on the unit circle, because the problem requires a 50 Ohm load in the terminating network (the oscillator output). There are many values that can be chosen to make port 1 unstable. To avoid many different solutions, a convenient value is Gamma_T = 0.7 at an angle of 135 degrees. This is on the 1+jb circle, which means the terminating network can be a stub and a 50 Ohm load.
This value of Gamma_T on port 2 makes Gamma_in looking into port 1 unstable. Find Gamma_in and the corresponding value of Z_in = -R + jX. The tuning network attached to port 1 should have an impedance of ZL = R/3 - jX. Design a network to realize this value with a length of line and resistor. Use the a length of line to rotate this impedance towards the load onto the real axis, then read off the value of the resistance from the real axis of the Smith chart.
For the terminating network on port 2, use the method for maximizing instability from the lecture notes to find Gamma_in_max. From Gamma_in_max, find the value of Gamma_T needed from the network attached to port 2 to achieve this value of Gamma_in. Unlike the previous problem, this will be a purely reactive value, so the terminating network can be realized with just an open circuit stub. Design its length using the Smith Chart.
For the tuning network, find the value of Z_in = -R+jX corresponding to Gamma_in_max, and as before the tuning network should have an impedance of ZL = R/3 - jX. Design a network to match a 50 Ohm load to this impedance using a length of line and an open circuit stub.