(i) Using $n^a \times n^b = n^{a+b}$: $2^{-4} \times 2^7 = 2^{-4+7} = 2^3$ (ii) Using $n^a \times n^b = n^{a+b}$: $3^2 \times 3^{-5} \times 3^6 = 3^{2-5+6} = 3^3$ (iii) Using $n^a \times n^b = n^{a+b}$: $p^3 \times p^{-10} = p^{3-10} = p^{-7}$ (iv) First, $(-4)^{-2} = \frac{1}{(-4)^2} = \frac{1}{16}$. Then $2^4 \times \frac{1}{16} = 16 \times \frac{1}{16} = 1 = 2^0$ (v) Using $n^a \times n^b = n^{a+b}$: $8^p \times 8^q = 8^{p+q}$