The answer lies in the 3rd inequality
\(a > -b\)
i.e. \(a + b > 0\)

We can go for option C

Alternatively:

\(ab < 0\)
So, wither \(a\) is +ve and \(b\) -ve OR \(a\) is -ve and \(b\) +ve

\(a > b\)
This means \(a\) cannot be -ve
So we stick to \(a\) is +ve and \(b\) -ve

\(a > -b\)
\(-b\) would be a +ve number and \(a\) is more +ve than it

Let us assume two number which satisfy all three inequalities
\(a = 3\) and \(b = -2\)

A. \(\frac{3}{-2} > 0\) - NO

B. 3 + (-2) < 0 - NO

C. (-2) - (-3) > 0 - YES

D. \(\frac{3}{-2} = 1\) - NO

E. (3) - (-2) < 0 - NO