The following is an exercise in Just/Weese (page 179),
Question 1: can you tell me if I got it right? Thank you!
Question 2: Shouldn't it be equality rather than less equals in $\mu = \sum_{\alpha < \kappa} |\alpha|^\lambda \color{red}{\le} |\alpha_0|^\lambda \cdot \kappa$?
Assume $|\alpha_0|^\lambda < \kappa$. Then $|\alpha|^\lambda < \kappa$ for all $\alpha < \kappa$. Then $\displaystyle \sum_{\alpha < \kappa}|\alpha|^\lambda = |\alpha_0|^\lambda \cdot \kappa = \kappa$.
We know that if $\kappa$ is an infinite cardinal then $\kappa$ is singular if and only if there exists an $\alpha < \kappa$ and a set of cardinals $\{ \kappa_\xi : \xi < \alpha \}$ such that $\kappa_\xi < \kappa$ for all $\xi < \alpha$ and $\kappa = \sum_{\xi < \alpha} \kappa_\xi$.
Hence if $|\alpha_0|^\lambda < \kappa$, then $\kappa$ must be regular since there is no $\gamma < \kappa$ with $\sum_{\alpha < \gamma} |\alpha|^\lambda = |\alpha_0|^\lambda \cdot \gamma = \kappa$. But by assumption, $\kappa$ is singular.