Answer
$A\cup B^{\prime}=\{a,b,d\}$
$A\times B^{\prime}=\{(a,a), (a,d), (b,a), (b,d)\}$
Work Step by Step
If $S$ is the universal set and $A\subseteq S$,
then $A^{\prime}$ is the complement of A (in $S$),
the set of all elements of $S$ not in $A$.
$A^{\prime}=\{x\in S|x\not\in A\}$
For an element to be in $A\cup B$, it must be in $A$ or in $B$.
For an element to be in $A\cap B$, it must be in $A$ and in $B$.
A$\times B=\{(a, b)|a\in$A and $b\in B\}$
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$B^{\prime}=\{a,d\}$
$A\cup B^{\prime}=\{a,b,d\}$
$A=\{a,b\}, B^{\prime}=\{a,d\},$
$A\times B^{\prime}=\{(a,a), (a,d), (b,a), (b,d)\}$
