It’s not really any different from proving the finite versions: just show that each of the sets is a subset of the other. For (i), for instance, you want to show that
$\left(\bigcup_{n\ge 1}A_n\right)'\subseteq\bigcap_{n\ge 1}A_n'\tag{1}$ and that
$\bigcap_{n\ge 1}A_n'\subseteq\left(\bigcup_{n\ge 1}A_n\right)'\;.\tag{2}$
To show $(1)$, assume that $x\in\left(\bigcup_{n\ge 1}A_n\right)'$; then $x\notin\bigcup_{n\ge 1}A_n$. This means that for every $n\ge 1$, $x\notin A_n$, which by the definition of complement means that $x\in A_n'$ for every $n\ge 1$. But that’s exactly what it means to say that $x\in\bigcap_{n\ge 1}A_n'$, so I’ve just proved $(1)$.
To prove $(2)$, assume that $x\in\bigcap_{n\ge 1}A_n'$. From the definition of intersection this means that $x\in A_n'$ for every $n\ge 1$, and hence that $x\notin A_n$ for every $n\ge 1$. This in turn means that $x\notin\bigcup_{n\ge 1}A_n$, i.e., that $x\in\left(\bigcup_{n\ge 1}A_n\right)'$, so $(2)$ is also true. Finally the truth of $(1)$ and $(2)$ ensures that
$\left(\bigcup_{n\ge 1}A_n\right)'=\bigcap_{n\ge 1}A_n'\;.$
I’ll leave (ii) to you; you should be able to use much of what I did here as a model.