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

To solve the first equation, \(2^{x} = \frac{1}{32}\), we can rewrite \(\frac{1}{32}\) as \(2^{-5}\) because \(32 = 2^5\). This gives us the equation \(2^{x} = 2^{-5}\). Since the bases are the same, we can set the exponents equal to each other, so \(x = -5\). For the second equation, \(8^{a+1} = 1\), we recognize that anything raised to the power of zero equals 1. So, we can write \(a+1 = 0\). Solving for \(a\), we find \(a = -1\). Thus, the solutions are \(x = -5\) and \(a = -1\).