1) 2x-3y = -2
2) x>2 and y>0
this is a good one.
from i, 2x-3y = -2. this is sufficient because the left part of the inequality is negative and the right part is 0 or greater than 0 because square of any value cannot be negative.
from ii, x>2 and y>0. this is also sufficient because x and y could have any values but 2x-3y is always smaller than x^2. in only case 2x-3y would be greater than x^2 is when y is negative but it clearly is not. so this is also sufficient.
if needed, will explain more....