\( \begin{array}{ll}\text { 2.3. } & \text { Solve the following equations: } \\ \text { 2.3.1. } & 2^{x}=\overline{32} \\ \text { 2.3.2. } & 8^{a+1}=1\end{array} \)

To solve the equation \( 2^{x}=\overline{32} \), we recognize that \( \overline{32} \) represents the number 32. Since \( 32 = 2^5 \), we can set the exponents equal: \( x = 5 \). For the second equation \( 8^{a+1}=1 \), we can rewrite 8 as \( 2^3 \), so we have \( (2^3)^{a+1} = 1 \). Since any number raised to the power of 0 equals 1, we set \( 3(a+1) = 0 \), giving us \( a + 1 = 0 \) or \( a = -1 \).