(1) (−4)0=1(-4)^0 = 1(−4)0=1 (2) (−2)−3=1(−2)3=−18(-2)^{-3} = \frac{1}{(-2)^3} = -\frac{1}{8}(−2)−3=(−2)31=−81 (3) 32⋅3−252−50=125−1=124\frac{3^2 \cdot 3^{-2}}{5^2 - 5^0} = \frac{1}{25 - 1} = \frac{1}{24}52−5032⋅3−2=25−11=241 (4) (2+2−1)−1=(2+12)−1=(52)−1=25(2 + 2^{-1})^{-1} = \left(2 + \frac{1}{2}\right)^{-1} = \left(\frac{5}{2}\right)^{-1} = \frac{2}{5}(2+2−1)−1=(2+21)−1=(25)−1=52 (5) (3+3−1)−2=(3+13)−2=(103)−2=9100(3 + 3^{-1})^{-2} = \left(3 + \frac{1}{3}\right)^{-2} = \left(\frac{10}{3}\right)^{-2} = \frac{9}{100}(3+3−1)−2=(3+31)−2=(310)−2=1009