To determine whether the expression \( (0.5)^n + (0.5)^{-n} \) is always greater than 2 for an integer \( n \), let's analyze the behavior under different conditions. Consider the expression:

\(a = (0.5)^n + (0.5)^{-n}\)

We must examine whether \( a > 2 \) under the given statements:

• Statement I: \( n \) is an even integer
• Statement II: \( n \) is negative

Let's analyze each statement:

1. Statement I: \( n \) is an even integer
   • If \( n \) is an even integer, then \( n = 2k \) for some integer \( k \).
   • Substituting, we have \( (0.5)^n = (0.5)^{2k} = (0.5^2)^k = (0.25)^k \) and \( (0.5)^{-n} = (0.5)^{-2k} = (2^2)^k = 4^k \).
   • Thus, \( a = (0.25)^k + 4^k \).
   • It's evident \( 4^k \geq 1 \) for \( k \geq 0 \), and \( (0.25)^k \leq 1 \), so \( a \) may not always be greater than 2 for non-negative even integers.
   • For \( k < 0 \), \( (0.25)^k > 1 \), and \( 4^k < 1 \), so \( a = (0.25)^k + 4^k\) may or may not exceed 2 depending on \( k \).
2. Statement II: \( n \) is negative
   • If \( n \) is negative, then \( (0.5)^n = 2^{|n|} \) and \( (0.5)^{-n} = 2^{-|n|} \).
   • So, \( a = 2^{|n|} + 2^{-|n|} \).
   • For \( n < 0 \), \( |n| > 0 \), thus \( 2^{|n|} \geq 1 \) and \( 2^{-|n|} < 1 \).
   • Since \( f(x) = 2^x + 2^{-x} \) is always greater than 2 for any non-zero \( x \), \( a > 2 \) for all negative \( n \).

Based on our analysis:

• Statement II is sufficient to determine that the expression is always greater than 2 for negative integers.
• Statement I is not sufficient by itself because it doesn't ensure that the expression exceeds 2 for all even integers.

Hence, the correct answer is: The Question can be answered by using one of the statements alone, but cannot be answered using the other statement alone.