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Bunuel
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is a>b,

a & b are consecutive even number ie. (a,b) = (-2,0), (0,-2), (2,4), (4,2)... and so on.
statement 1 : (a+2) & (b-2) are consecutive integers. which shows that a<b and hence is sufficient to answer no that a is not > b
Statement 2 : (b-3) & (a+1) are consecutive odd integers. which again after plugging different values give that a<b and hence sufficient to answer that a is not > b.
hence answer is D as both statements are sufficient to answer that a<b.
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In the case of

a:0 b:2 & a:-2 b:0 both satisfy statement 1, how can D be correct?

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So ur statement gives that a<b as well and hence answers the question that a is not >b

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If a and b are consecutive even integers, is a > b?

(1) a + 2 and b - 2 are consecutive even integers
I: a > b
a = b+2
\(\implies\) b+2+2 and b - 2 are consecutive even integers which is not possible.

II: a < b
b = a + 2
\(\implies\) a+2 and a are consecutive even integers which is TRUE.

SUFFICIENT.

(2) b - 3 and a + 1 are consecutive odd integers
I: a > b
a = b+2
\(\implies\) b - 3 and b + 3 are consecutive even integers which is not possible.

II: a < b
b = a + 2
\(\implies\) a-1 and a+1 are consecutive even integers which is TRUE.

SUFFICIENT.

Answer D.
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