1. \( 2^{x+1}=256 \) 2. \( \frac{a^{2} b^{3}}{a c^{2}} \times \frac{4 a^{-2} b c}{20 b^{3}} \) 3. \( \left(3^{-1}\right)^{-1} \)

To solve the first equation, rewrite \( 256 \) as \( 2^8 \). This gives you \( 2^{x+1} = 2^8 \), which means \( x + 1 = 8 \). Solving for \( x \) gives you \( x = 7 \). Easy peasy, right? For the second expression, simplify step by step. Start by reducing: \( \frac{a^{2} b^{3}}{a c^{2}} = \frac{a^{1} b^{3}}{c^{2}} \) and \( \frac{4 a^{-2} b c}{20 b^{3}} = \frac{a^{-2} c}{5 b^{2}} \). Combining gives: \( \frac{a^1 b^3 \times a^{-2} c}{c^2 \times 5 b^2} = \frac{4 a^{-1} b^1}{5 c^{1}} \). A little bit of algebra magic, and you have your answer!