A. in order to keep the cards, the sum must be > 0. If the first card is negative the second card must be > the negative of of the same card.
example: 0.3 + -0.2 = 0.1
-0.2 is > -0.3
B. The card drawn is a -0.2. Same solution as (A) the second card must be > the negative of the card drawn. So she must draw > 0.2.
(0.5, 0.4, 0.3)
C. The card drawn is a -0.2. If the cards have to be put back due to negative sum, he drew a second card <0.2. This solution is the reverse of (B) except you discount 0.2 because this give a zero sum, not negative. (-0.5, -0.4, -0.3, -0.2, -0.1, 0.1).