This doesn't seem right to me. A capacitor doesn't automatically fail once it reaches its rated voltage. Your 10V capacitor may continue to function just fine as a capacitor, at 11V or 12V. The rated voltage is a bit like the best before date on a litre of milk. Don't trust it past this voltage, but don't expect immediate failure.

Yes, without a doubt, this is currently the best answer here.

Oh heavens no! Please do not suggest that someone can short a battery with a wrench for up to 30 seconds. You're off by a factor of hundreds, and by 30 seconds, you are likely to have melted the wrench, exploded the battery, suffered severe burns, sung with the angels, or all of the above. An answer like this is dangerous.

This seems incorrect to me.

This is the right way of looking at this. The teacher is correct in the LIMIT where R1 and R2 are much greater than R3, but has messed up by not showing R1 and R2 on the diagram. The OP is correct in the limit where R3 is much greater than R1 and R2. The real-world result is going to be somewhere in between.

@IanBland I can just picture the scenario when an electrician has to turn the van around because they've left their gerbils at home.

As you plan this out, draw your intended circuit on paper first, and calculate the voltage across each LED to make sure it's in the correct range. It may actually be possible to do this without adding resistors to the circuit, which would be the most efficient type of circuit.

You didn't paste the information into the answer. Only the link. Stack Exchange sites generally prefer to have the actual content in the answer, in case the link breaks in the future. We don't need the whole lecture - just the bit that explains what complex values for V and I actually mean, or a summary thereof.

The answer itself doesn't really answer the question. But the text in the link in the previous comment does. Perhaps, @EugeneSh., you could paste the relevant information into the actual answer, to make it a complete answer by itself.