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Answer :
(i) We have:
\(3^{-2}=\frac{1}{3^2}=\frac{1}9\quad\)[∵\(a^{-n}=\frac{1}{a^n}\)]
(ii)We have:
\((-4)^{-2}=\frac{1}{(-4)^2}=\frac{1}{16}\quad\)[∵\(a^{-n}=\frac{1}{a^n}\)]
(iii)We have:
\((\frac { 1 }{ 2 })^{-5}=2^5=32\quad\)[∵\((\frac{a}b)^{-n}=(\frac{b}a)^n\)]