The question is asking that given x>0, is x^2<x?

x^2<x means x^2-x<0
Or x(x-1)<0
So either (a) x is positive and (x-1) is negative
or
(b) x is negative and (x-1) is positive.
(a) implies x>0 and (x-1)<0 which is 0<X<1.
(b) implies x<0 and (x-1)>0 which is x<0 and x>1.
Case (b) is automatically rejected because a value cannot be both less than zero and more than 1.
So the question becomes is 0<x<1?
Consider first statement (1) alone.
0.1<x<0.4. Or 0<x<1.
So answer to the question is yes.
Or statement (1) alone is sufficient to answer the question.
We next consider statement (2) alone.
x^3<x^2 .
Or (x^3-x^2)<0
Or x^2(x-1)<0
Since x^2 is always positive, (x-1)<0 which is x<1.
Already main question says that x>0.
So we have that 0<x<1 from statement (2) and the answer to the question is yes.
Or statement (2) alone is also sufficient.
The correct answer is hence (D).