zero vector means nonempty how do you know?
how do you know if the subset has a zero vector or not...
example
W = { (x1,x2,1): x1 and x2 are real numbers }
solution: because 0 = (0,0,0) is not in W, you know that W is not a subspace of R3.
example 2: W = { (X1, X1+x3, X1) : x1 and x3 are real numbers}
solution: this set is nonempty because it contains the zero vector (0,0,0).
WHERE ARE THEY FIND THIS ZERO VECTOR, OR LACK THERE OF!?!?! PLEEEEASE HELP. SOO CONFUSED. 
Comments
Well, for the first, if (0,0,0) were in there, then we'd have
(x1, x2, 1) = (0, 0, 0)
which is actually three equations:
x1 = 0
x2 = 0
1 = 0
The first two equations are fine; however, the last, 1=0, is a contradiction, so (0,0,0) must not be in our W.
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Do similarly for the second:
If (0,0,0) is in W, then we have (x1, x1+x3, x1) = (0,0,0).
x1 = 0
x1 + x3 = 0
x1 = 0
Solving, we get x1 = 0 and x3 = 0.
These are real numbers, so this W does contain a zero vector.
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Hopefully that's what your question was.
Ok. In example 1, you have a set of vectors with the form (x1, x2, 1), with x1 and x2 being real numbers.
This means in that set, you have the vectors (4, 5, 1), (2, -2, 1), (0, 0, 1), (-2, -123123, 1), whatever.. as long as the last coefficient in the vector is 1, then it can exist in that set.
The problem in that case is that the zero vector (0, 0, 0) doesn't exist in that set because the last coefficient must be 1.
It's like someone asking you: "If you have a vector space where the last coefficient of all the vectors in there is 1, is the vector with three zero coefficients in there?" The answer is obviously no.
Second example:
W = { (X1, X1+x3, X1) : x1 and x3 are real numbers}
Well, you're asking if this set is empty. Why not test the most trivial case? Set x1 and x3 to zero (which is a real number).
You get - tada - the zero vector. Which exists in that space, because it satisfies the (x1, x1+x3, x1) criteria. So the set's clearly not empty.
It takes a little bit of a mental leap but try to think about it logically.