in #1 you have considered e as a fraction....
Bunuel
If e is a nonzero integer, is 1/2^e greater than or less than 1?
(1) −e < 1
(2) e^2 > 0
Ok (1/2) ^ e, where e does not equal 0 and e is an integer will be greater than 1 if e <0 (a fraction to a negative exponent gives us 1 divided by some number less than 1, and thus we get a number greater than 1). So the question simplifies to is e <0 Yes/No?
(1) -e <1 gives us e >-1 Let
e =-1/2. Then (1/2)^(-1/2) = 1/((1/2)^1/2) >1 so we get a YES. Let e =2, then (1/2)^2 = 1/4 and we get a NO. Since we get a Yes and a No, we can mark NS
(2) e^2 >0. Therefore either e>0 or e <0. Let e =-1, then we get a YES answer (since (1/2)^-1 = 2). Let e = 1. then we get (1/2)^1 = 1/2 and we get a No answer NS
(1) and (2) e>-1 and either e>0 or e<0. First lets consider e >-1 subject to e>0. Suppose e = 1. then we get a No answer (since 1/2^1 = 1/2). Now lets consider e >-1 subject to e<0. This gives -1<e<0. Let e = -1/2. Then We get a Yes answer (since 1/2^(-1/2) = 1/((1/2)^1/2) > 0). Since we can get a Yes answer from one case and a no answer from the other case, statements (1) and (2) combined are Not Sufficient.
The answer is E.