In this case answer is equal to 2^ones(w) where ones(w) is amount of ones in binary representation of w.

a&b = 0 says that there is no common set bit in a and b. In this case sum operation equals to bitewise-or operation, because there is no digit overflows in our sum.

So now we have these equations:

a&b = 0

a|b = w

As we only use bitwise operations we can solve the problem for each bit independently and then multiply our results for all bits.

If current bit in w is zero, only possible solution is a = 0; b = 0. Answer is 1.

If current bit in w is one, there are two solutions: a = 0; b = 1 and a = 1; b = 0. Answer is 2.

Final answer is 2^ones(w)·1^zeroes(w) which is equal to 2^ones(w).

If w = 0 then we have a = b, and the only possible solution for this is a = b = a&b = 0.

If w is positive we can generate infinte amount of pairs (a, b) that satisfy both equations. Let's onsider w = 13 and look at binary presetation of some correct pairs (a, b). It is easy to see the pattern that generates some of them:

w |    1101
a | +    11
b | = 10000

w |     1101
a | +  10011
b | = 100000

w |      1101
a | +  110011
b | = 1000000

w |       1101
a | +  1110011
b | = 10000000

w |        1101
a | +  11110011
b | = 100000000

w |         1101
a | +  111110011
b | = 1000000000

As we have no other limitations answer is ∞ for any positive w.