Second address is technically incorrect. The loopback address subnet 127/8 does not contain private network addresses like 192.168/16, 172.16/12, or 10/8 and doesn’t provide any utility for routing within the local network.

Standard resistors come in many standard values and many power grades, meaning you can pick a resistor that minimizes source current for your application. Incandescent bulbs come in different rated voltages and powers, but they are far from precise or standard.

Depending on the design, shunt regulators come with short protection for free without relying on a PTC fuse (incandescent bulb or otherwise). You’re absolutely correct that the positive tempco of a bulb will reduce short circuit current, but how by how much and whether it is worth it depends on the application and the properties of the bulb itself.

The lamp intensity won’t vary in a shunt regulator because the current through the shunt resistor is constant. The current through the zener diode varies to split that current with the load. Otherwise, yes it will work, however you will have fewer choices of resistance and will not be able to make the regulator as efficient.

I use Altium Designer and KiCAD. KiCAD isn’t as good by just a hair but makes up for it by being free and open-source. Altium Designer is also crazy expensive for hobbyists.

Yes you can absolutely breadboard it. Forcing a current is as easy as following ohm’s law. Make sure there is a certain voltage across a resistor and ensure that only a negligible amount of that current is leaked elsewhere. A difference amplifier is a good way to ensure this, as long as you pay attention to the amplifier input currents.

If current regulation isn’t super important, a highish voltage (say 24+ V) and a large resistor will also work because the variation of threshold voltage will be so small that the voltage across the resistor will be relatively stable.

I think there is some confusion about the word diode here. The transistor is effectively an inverting amplifier, that is that the drain voltage is reduced if the gate voltage is increased. By tying them together, they reach a stable configuration where the gate is just high enough to make the drain low enough for them to be equal. In this configuration, there are two terminals, hence the di in diode. Like a traditional diode, it has a very nonlinear voltage-current relationship. If you apply 10V to it, theoretically the current would be thousands of amperes. Practically that won’t happen but you will blow up the transistor.

I don’t know enough about radiation and semiconductor physics to answer your other questions but if I were you I would just build it and test it. MOSFETs and resistors are cheap and if you do have a radiating source on hand it might be easier to try and fail than to hope someone here can tell you what your part will do when exposed to conditions outside of the manufacturer recommendations.