also thus the ignited precipitate mean that its free of H20? (ignited precipitate=CaC2O4?).

Yes. the idea behind the baking/burning of the precipitate is to drive off any water and CO2 from the precipitate to convert CaC2O4*H2O to CaO.

The calculation Breizh did, as he noted, is related to the purity of the sample. Sometimes I've seen this called a % recovery calculation. You compute the quantity of CaO that the reaction should produce assuming the original CaCO3 is 100% pure and none of the Ca is lost over the course of the analysis (Breizh showed 0.85 g). Compare to the actual amount of CaO recovered, and you get 97% recovered. This number means that, either the original CaCO3 was not 100% pure (actually about 97%), or some of the CaO was lost over the course of the analysis.

but I think that the %CaO in the sample should be calculated as (weight of CaO in the sample)/(weight of sample)?

I think you are correct, that this is the kind of percentage they are looking for, but I'm not sure how they got 24%, because mass CaO/mass CaCO3 is bout 54%. Theoretically, CaCO3 (CaO*CO2) should be 56% CaO (weight of CaO/weight of CaCO3). CaO would be 38-39% of the hydrated precipitate.

I'm missing something, I'm sorry. Maybe these ramblings will help in some way, though.