Answer
a) $\frac{16}{3}$
b)$1$
Work Step by Step
a) $\frac{4\cdot3^{-2}}{2^{-2}\cdot3^{-1}}=\frac{4\cdot2^2\cdot3^{1}}{3^2}=\frac{4\cdot2\cdot2\cdot3}{3\cdot3}=\frac{4\cdot2\cdot2}{3}=\frac{16}{3}$
b)$(-2)^0=1$
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