X+Y can take the values 0, 1, 2, with probability 1/4, 1/2, 1/4

|X-Y| can take the values 0, 1, with probabilty 1/2, 1/2

Also the RV ((X+Y), |X-Y|) can take values (0,0), (0,1), (1,0), (1,1), (2,0), (2,1),

with probabilities 1/4, 0, 0, 1/2, 1/4, 0.

So now you can just evaluate E((X+Y)(|X-Y|)) and E(X+Y)E(|X+Y|) to show that

they are equal

If X+Y and |X-Y| are independent then their joint distribution:

P((X+Y)=a, |X-Y|=b)=P((X+Y)=a)P(|X-Y|=b)

and you have sufficient information to check this

RonL