1. Stating the problem: Given $a = 2^5 + 2^{-5}$ and $b = 2^5 - 2^{-5}$, find $a^2 - b^2$. 2. Recall the identity: $$a^2 - b^2 = (a-b)(a+b)$$ 3. Calculate $a+b$ and $a-b$: $$a+b = (2^5 + 2^{-5}) + (2^5 - 2^{-5}) = 2^5 + 2^{-5} + 2^5 - 2^{-5} = 2 \times 2^5 = 2^6 = 64$$ $$a-b = (2^5 + 2^{-5}) - (2^5 - 2^{-5}) = 2^5 + 2^{-5} - 2^5 + 2^{-5} = 2 \times 2^{-5} = 2^{-4} = \frac{1}{16}$$ 4. Now multiply: $$a^2 - b^2 = (a+b)(a-b) = 64 \times \frac{1}{16} = 4$$ 5. Final answer: $a^2 - b^2 = 4$ which corresponds to option E. --- Since the user asked two questions, but per instructions only the first is solved here.