Thanks Sayzer,

Am I right to calculate the capacitor's capacity like this (sorry but I don' t know the english technical names):

Assuming a very bad case where my PIC consumes 1mA and I would like my PIC to stay powered 2 hours, it would give this:
Q(cap load) = I(amps) x t(seconds)
this would give: 1mA x 7200seconds = 7,2C(Coulomb - capacity)

Then, to find out the cap's capacity, I would calculate the time like this:
C(Capacity) = Q(cap load) / U(Volt)
this would give: 7,2C / 5V = 1,44F

This is theory but please let me know if I'm wrong and if yes, where.

What about the low power programming?