In the Experiment 2 you found that with a current of Iout = 0.5 A not only the load restistor, but also the voltage regulator heat up significantly.
This is not a threat to the voltage regulator 78xx since it has an overheating protection and shuts down. However, this can cause all sorts of problems with your circuit when a voltage regulator to shut down unexpectedly. You should understand how the heat is generated and how to take care of it.
The power dissipation PD of the 78xx, and thus the generation of heat, is given by the formula...
PD = (Uin - Uout) × Iout + Uin × IGND
You can neglect the part (Uin × IGND) as IGND is only a few μA. Therefore, the calculated power dissipation...
PD = (Uin - Uout) × Iout.
= (9.0 V - 5.0 V) × 0.50A = 2.0 Watt.
The resulting heat must be dissipated and you might need a heat sink to dissipate enough heat, for the voltage regulator to remain functional.
However, you can also try to limit the power dissipation...