Simple diode question
I suspect your confusion comes from an incomplete picture of what a packaged diode is compared to the p-n junction diode you have learned about. You might notice that a discrete diode has metal leads coming out from the package, but your p-n junction diode model only has semiconductor. In order to interface with semiconductors we usually need to connect metal wires to them. Every introductory semiconductor book I have seen has a chapter on metal-semiconductor junctions which you can read about if you are interested. The short version is that the metal-semiconductor junctions also have a built in potential that cancels out the p-n junction built in potential. So to quickly address each of your specific questions:

1) If you short a diode (in the dark) you will have 0 current through everything, KCL is satisfied. The KVL equation looks something like \$0 = V_{bi,m1} + V_{bi,pn} + V_{bi,m2}\$ with one or two of those variables being negative which allows KVL to be satisfied.

2) If you forward bias a diode the built in voltage increases by the applied bias... sort of. This is technically going to be spread across all 3 junctions, but the vast majority should show up across the semiconductor p-n junction, so its fine to assume thats where it all is. KVL equation: \$-V_{ext} = V_{bi,m1} + (V_{bi,pn} - V_{ext}) + V_{bi,m2}\$

3) Basically the same as #2 but with a positive voltage: \$V_{ext} = V_{bi,m1} + (V_{bi,pn} + V_{ext}) + V_{bi,m2}\$