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(1) \(x>y+5\) --> \(x-y>5\). Clearly insufficient, for example: if \(x=1\) and \(y=-10\) then the answer is NO, but if \(x=10\) and \(y=1\) then the answer is YES. Two different answers, hence not sufficient.

(2) \(x^2-y^2=0\) --> \((x-y)(x+y)=0\) --> so either \(x-y=0\) or \(x+y=0\). Also insufficient: if \(x=1\) and \(y=1\), then answer is NO, buy if \(x=\frac{1}{2}\) and \(y=\frac{1}{2}\), then the answer is YES. Two different answers, hence not sufficient.

(1)+(2) As from (1) \(x-y>5\neq{0}\), then from (2) must be true that \(x+y=0\) --> so \(x=-y\) --> substitute \(x\) in (1) --> \(-y-y>5\) --> \(y<-\frac{5}{2}<0\), as \(x=-y\), then \(x>\frac{5}{2}>0\), so \(y^2\) (or which is the same \(x^2\)) will always be more than \(x\), thus the answer to the question "Is \(x>y^2\)" is NO. Sufficient.

To elaborate more as \(x=-y>0\), the only chance for \(x>y^2\) to hold true (or which is the same for \(x>x^2\) to hold true) would be if \(x\) is fraction (\(0<x<1\)). For example if \(x=\frac{1}{2}\) and \(y=-\frac{1}{2}\) then \(x=\frac{1}{2}>y^2=\frac{1}{4}\). But the fact that \(x>\frac{5}{2}>0\) rules out this option.

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