P(D l A) = 1/2 P(A)= 1/3 is completely wrong! What it means is that P(D|A)= 1/2 P(A) and that (1/2)P(A) is equal to 1/3- which made me think you intended P(A) to be 2/3! That's not true nor is it what you meant.
P(D|A)= 1/2 P(A) and P(A)= 1//3 is what you meant- do NOT connect things with "=" when they are not equal!
However, your real problem is that P(A) is NOT 1/3. That is the probability that a specific one of the players recovers. A is the event that one or the other recovers.
Call the two players "x" and "y". Let X be the event that player x recovers, Y the event that player y recovers. Then P(A)= P(X or Y)= P(X)+ P(Y)- P(X and Y). You are told that P(X)= P(Y)= 1/3 and the two recoveries are independent so P(A)= P(X or Y)= 1/3+ 1/3- 1/9= 2/3- 1/9= 5/9. The probabilty the one or the othe but NOT both, which you are counting separately is 5/9- 1/9= 4/9.
You could also argue that since the P(X)= P(Y)= 1/3, P(not X)= P(not Y), the probabilities that each does NOT recover, is 2/3. Then the probability that x recovers but not y is (1/3)(2/3)= 2/9 and the probability that y recovers but not x is (2/3)(1/3)= 2/9 and so the probability that one but not the other recovers is 2/9+ 2/9= 4/9.
Can Someone Please Cheak If My Answer Is Correct As I Am Not Really Sure.