It's a lot to expect, for a student to understand immediately how power factor behaves, how the formulas are derived and how to apply them.
A simulation can further our understanding, now that we have computers. I recommend a simple circuit consisting of an inductor and resistor, powered from 240VAC 60Hz. (Later you can add the correction capacitor.)
Examine voltage and current waveforms.
In particular notice how the Ampere waveform does not coincide with the supply voltage waveform, but is delayed for a fraction of a cycle.
For load impedance your approach agrees with the solution. 6.144 ohmic and some inductive. In series their combined effect produces power factor 0.8.
I guess 6kW is the 'real' power consumed. At 240V this calculates to 25A in 9.6 ohm effective impedance (6.144 + a few ohms 'choke' effect of the inductor).
Notice that power factor=.9 requires less inductor.
Power factor =.7 requires more inductor.
Your third image seems to omit some of your own calculations.
To select pfc capacitor value, the aim is to achieve resonance with the RL load at 60 Hz. Then the Ampere waveform shall coincide with the supply voltage waveform. The real Ampere level can be minimized. It drops to 25A.
It helps to experiment with a capacitive load too. You can observe the advanced timing of the Ampere waveform as compared to the supply voltage waveform. Experiment by trying various RC combinations.
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My above reply has to do with a single phase AC supply. It makes for a simpler learning environment. The exercise specifies a 3-phase supply which creates complications.