Yes, they are always true.
Take $x\in \bigcup_{i=1}^{\infty}\bigcup_{j=1}^{\infty}A_{ij}$. Hence there exists $i$ so that $x\in \bigcup_{j=1}^{\infty}A_{ij}$, and furthermore we find $j$ so that $x\in A_{ij}$. So for this $j$, $x\in \bigcup_{i=1}^{\infty} A_{ij}$, and hence $x\in\bigcup_{j=1}^{\infty}\bigcup_{i=1}^{\infty}A_{ij}$. With identical steps you can conclude the other inclusion as well.
For the second, take $x\in \bigcap_{i=1}^{\infty}\bigcap_{j=1}^{\infty}A_{ij}$. Hence for all $i$, $x\in\bigcap_{j=1}^{\infty}A_{ij}$, and thus for all $i$ and for all $j$, $x\in A_{ij}$. So for all $j$, $x\in\bigcap_{i=1}^{\infty}A_{ij}$, and thus $x\in\bigcap_{j=1}^{\infty}\bigcap_{i=1}^{\infty}A_{ij}$. The other inclusion is analogous.