Q1: I believe it is the at-infinity focal length. But it's not usually the easiest way to get the answer you want; just use the extension ratio as a multiplier. Say you're at 1:1 macro, the extension ratio is 2 (lens twice as far from film as when at infinity), therefore the effective f-number is twice what is marked on the barrel. If you compute exposure using the effective f-number instead of the marked f-number, you've performed the bellows-factor compensation.

Q2: yes.

Q3: time. Again, this is often not the easiest way to arrive at the answer you want because you probably cannot measure the image distance. For me at least, I generally start the bellows compensation calculation from magnification (M = image size / object size). And if you've computed exposure using effective f-number instead of marked f-number, you've already performed this correction.

bellows factor = (1+M)^2

Originally Posted by pbromaghin
it does appear that multiplying the Effective f-stop by the Exposure Factor does yield the Marked f-stop.
Nearly. The extension factor is E=M+1 and the effective focal length is f*E. The physical aperture does not change, therefore the effective f-number increases by a factor of E. Exposure goes as the square of f-number, hence the exposure correction factor of (M+1)^2