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.
Best Answer
The answers by @Vasiliy and @johnfound are incorrect. 1N400x diodes are not all identical except for "accidental" manufacturing variations.
Diodes with higher reverse voltage ratings are intentionally manufactured with lighter doping so that the depletion region for a given reverse voltage is wider than it otherwise would be. The disadvantage with lighter doping is that the forward resistance and voltage drop for a high-voltage diode is higher than it would be for a lower-voltage diode.
So, you can use a 1N4007 for all of your applications, but your circuit efficiency will be slightly higher if you use a more appropriately-rated diode in low-voltage applications.