x _ y = xy - y

1.) a _ b

That means ab - b. 

Let's set

ab - b = 0

b(a - 1) = 0

Using the zero factor property,

b = 0 or  a - 1 = 0
              a = 1


Since b is a positive integer we can rule out b = 0.
However if a = 1  then ab - b =0 when b is any positive integer. 

So 1.) can be zero.

-------------------------

2.) (a+b)_b

That means (a+b)b - b.  

Let's set

(a+b)b - b = 0

Factor out b:

b[(a+b)-1] = 0
b[a+b-1] = 0

Using the zero-factor property again:

b = 0;  a+b-1 = 0
            a = 1-b

We must rule out b = 0 since b must be a positive integer.
We also must rule out a = 1-b since a positive integer b subtracted
from 1 canNOT be a positive integer.

So (2) canNOT be zero.

----------------------------

3. a_(a+b)

That means a(a+b) - (a+b).  

Let's set:

a(a+b) - (a+b) = 0

Factor out (a+b)

(a+b)(a-1) = 0

Using the zero-factor property again:

a+b = 0 ;  a-1 = 0
  a = -b;    a = 1

We have to rule out a = -b because a and b are both positive integers,
but if a = 1, then

(1+b)(1-1) = (1+b)0 = 0, regardless of what positive integer b equals.

So 3.) can be zero

-----------------------

So only

2.) (a+b)_b

cannot be zero.  The other two can be zero.

Therefore the correct choice is

e.) one and three

Edwin