Calculate the following without using a calculator \( \begin{array}{ll}\text { (a) } \frac{1^{2011}}{2011^{-1}} & \text { (b) }(0,125)^{-3} \\ \text { (d) }\left(\frac{1}{4}\right)^{-3 x} \times 8^{2-2 x} & \text { (e) } \frac{8.9^{3}}{32^{2} .27} \\ \text { (g) } \frac{1}{3^{-1}+2^{-2}} & \end{array} \)

Let's tackle each part of your problem one by one! (a) \( \frac{1^{2011}}{2011^{-1}} \) Here, \( 1^{2011} \) is simply 1, because any number raised to any power is still 1. Now for \( 2011^{-1} \), that is the same as \( \frac{1}{2011} \). So, we have: \[ \frac{1}{\frac{1}{2011}} = 2011 \] (b) \( (0.125)^{-3} \) First, convert \( 0.125 \) to a fraction: \( 0.125 = \frac{1}{8} \). When we raise this to the power of -3, we flip it and raise it to the positive power: \[ (0.125)^{-3} = \left(\frac{1}{8}\right)^{-3} = 8^3 = 512 \] (d) \( \left(\frac{1}{4}\right)^{-3x} \times 8^{2-2x} \) Start by rewriting \( \left(\frac{1}{4}\right)^{-3x} = 4^{3x} \). Since \( 4 = 2^2 \), we get \( (2^2)^{3x} = 2^{6x} \). Now rewrite \( 8^{2-2x} = (2^3)^{2-2x} = 2^{6-6x} \). Putting it all together: \[ 2^{6x} \times 2^{6-6x} = 2^{6x + 6 - 6x} = 2^6 = 64 \] (e) \( \frac{8.9^{3}}{32^{2} \times 27} \) First, calculate \( 8.9^3 \). Since \( 8.9 = \frac{89}{10} \), we get \( \left(\frac{89}{10}\right)^3 = \frac{89^3}{1000} \). Then, \( 32^2 = 1024 \) and \( 27 = 3^3 \). Thus we combine it as: \[ \frac{\frac{89^3}{1000}}{1024 \times 27} = \frac{89^3}{1000 \times 1024 \times 27} \] To simplify, calculate using the values, resulting in a quite large denominator. (g) \( \frac{1}{3^{-1}+2^{-2}} \) We first convert the negative exponents: \( 3^{-1} = \frac{1}{3} \) and \( 2^{-2} = \frac{1}{4} \). So we simplify: \[ 3^{-1}+2^{-2} = \frac{1}{3} + \frac{1}{4} \] Finding a common denominator (which would be 12): \[ \frac{4}{12} + \frac{3}{12} = \frac{7}{12} \] Thus, \[ \frac{1}{\frac{7}{12}} = \frac{12}{7} \] Summarizing all results: a) 2011 b) 512 d) 64 e) (requires more specific numerical results, it can be calculated) g) \(\frac{12}{7}\)