1. Again, suppose Jeremy cuts the first cake into the fractions f and 1-f , where f is at least 1/2. Then if Marie chooses first, Jeremy will get 1 1/4 cakes from the last two. If Marie chooses to go second, then Jeremy will get f of the first cake and then will divide the last two cakes in half.

So, f + 1/2 + 1/2 = (1 - f) + 1 1/4.
That is, f + 1 = 2 1/4 - f, or 2f = 1 1/4.
That is f = 5/8. Therefore Jeremy will receive 1 5/8 cakes and Marie will receive 1 3/8 pieces.

2. There is an inductive pattern to these problems. If for k cakes where Marie gets the first choice in all but one, Jeremy still has an advantage A (that is, Jeremy would receive A more cake than Marie among those k), then Jeremy will still have an advantage for k+1 cakes.

Here is why. Jeremy, knowing that he will have an advantage of A if Marie chooses first for the first cake, cuts the first piece into two pieces, 1/2 + A/4 and 1/2 - A/4. If Marie, chooses the piece having 1/2 + A/4, then Jeremy will suffer a disadvantage of A/2 for that first cake, but will gain an advantage of A for the remaining k cakes, so have a net advantage of A/2. By contrast, if Marie chooses to go second for the first cake, then Jeremy will cut the remaining cakes in half. Again he will have an overall advantage of A/2. Thus, Jeremy's advantages for from 1/2 for two cakes, to 1/4 for three cakes, to 1/8 for four cakes, to ... 1/64 for seven cakes.

3. Yes, simply make it so that Marie gets the first choice every time. Jeremy will then be forced to cut evenly every time.