W1 is not a subspace of R2:
Example: (3,0) and (0,3) are both in W1. But (3,0) + (0,3) = (3,3) is not in W1.
W2 is a subspace of R2:
Proof:
(1). Let (x, y) and (a, b) be in W2. Then, x+y = 0 and a+b = 0.
So, (x,y) + (a,b) = (x+a, y+b) which is in W2 since (x+a) + (y+b) = (x+y) + (a+b) = 0.
Therefore, W2 is closed under addition.
(2) Let (x,y) be in W2 and let c be a real number. So, x + y = 0.
c(x,y) = (cx, cy). cx + cy = c(x+y) = c(0) = 0. Thus, c(x,y) is in W2.
Therefore, W2 is closed under scalar multiplication.
Since W2 is closed under both addition and scalar multiplication, W2 is a subspace of R2.