Electronic – Energy subtracted from a capacitor for a given voltage drop

You have wrong calculations: In the first case, the usable energy is 10.88J; In the second case - 12.18J. (The full energy of the capacitor charged to 5V is 12.5J) $$ E_{start} = U_0^2.C/2 $$ $$ E_{end} = U_1^2.C/2 $$ $$ E_{usable} = E_{start} - E_{end} = (U_0^2 - U_1^2).C/2 $$

So, the difference is not so big. One more example, if you use only buck regulator discharging the capacitor to 2.5V the energy will be 9.375J.

So, the efficiency can be very very important in this case. If you have one regulator with 85% efficiency, the final gain is 10.88J*0.85 = 9.248J

But if you have two stages by 85% you will have 12.18J*0.85*0.85 = 8.8J

So, to resume: In order to use two stage regulator you need every stage to have efficiency higher than 90%.

Even using one stage, buck regulator and discharging to 2.5V can be a good solution if you have regulator with high enough efficiency.

Notice for low power: As a rule, the efficiency of the regulators degrades on low powers. The general purpose regulators are usually not optimized for extremely low output currents. (Although 1mA is still not so small current, so standard solution still can exists).

If the needed current is very small (for example 5uA), I would suggest designing of your own switching regulator, where the regulator consumption could be made small enough in order to not affect the efficiency. Check "The Art of electronics" - there is a good example of micro power design.

Considering the top circuit :

Applying KCL at the nodes :

A : (Vd - Va) / R1 = ( Va / sC1) + ( (Va - Vb) / sCx)

B : ( (Va - Vb) / sCx ) + ( (Vd - Vb) / sC2) = ( Vb - VL) / R2

L : ( (Vb - VL) / R2 ) = VL / sCL

3 equations for 3 unknowns.

( Don't forget to put Laplace transform of input ie of Vd).

(Did I miss something ?)