Electronic – What happens if I put the DC voltage source AFTER the diode instead of before

My, that's a lot of questions.

Presumably the diode model you instructor wants you to use is an ideal diode with 0.7V drop in series with a 4 ohm resistor. So, when the diode is conducting, it behaves like a voltage source with a resistor in series. When the forward bias is less than 0.7V it does not conduct.

Is the 4 ohms per diode negligible in comparison to 1.5K? Well, it's more than 0.5% for two diodes so it will drop tens of mV. That might be negligible or it might not be, depending on the application. Since the instructor gives you the value, I suggest it might not be negligible in terms of getting the correct answer.

Two such diodes in series behave like one 1.4V diode with 8 ohms in series.

Answering your second question first...

Heavier-doped junctions have less leakage current. Heavier-doped junctions also have a thinner depletion width, so they have more capacitance. So taking both measurements seems redundant.

What breaks the redundancy is when you realize that other factors that can contribute to leakage and capacitance.

Majority carrier diodes (i.e. Schottky diodes) are constructed by interfacing a metal with a semiconductor. The interface introduces crystal defects and charge traps. This causes increased leakage which doesn't correlate well with the capacitance. The defect density will depend on processing nuances, so you can't expect it to remain perfectly the same throughout the product's lifetime.

Capacitance is also introduced by interconnect. Especially for lateral devices, anode and cathode interconnect can be interleaved quite tightly to permit high currents to flow under forward bias, at the cost of a fixed additional capacitance.

To summarize, while in theory the measurements of leakage and capacitance are redundant, in practice they are not.

Answering your first question...

You're right to say that measurements of \$C\$ become less accurate when \$2\pi fC<G\$. However, the accuracy of such a measurement really depends on your test instrument.

The operation of an impedance analyzer is similar to a lock-in amplifier. An AC stimulus is applied, and two AC measurements are taken: one in-phase measurement, and one out-of-phase measurement. The in-phase measurement tells \$G\$, and the out-of-phase measurement tells \$B\$ (susceptance, from which \$C\$ is derived).

The ability of the instrument to resolve \$B\$ from \$G\$, i.e. to tell the difference between the in-phase component and the out-of-phase component, depends strongly on the details of the instrument's design. It's quite possible to get a reasonably accurate measure of \$B\$ when \$G\$ is 10, 100, or even 1000 times larger. But again, it depends on the instrument design.

The best thing to do is to refer to the instrument's user manual.