(A) The sample space $S$ for throwing two dice consists of $36$ outcomes.
Event $A$ is getting an even number on the first die:
$A = \{(2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$
Event $B$ is getting an odd number on the first die:
$B = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)\}$
The complement of $B$,denoted as $B^{\prime}$,consists of all outcomes in $S$ that are not in $B$. Since the first die can only show an even or odd number,the complement of getting an odd number on the first die is getting an even number on the first die.
Therefore,$B^{\prime} = \{(2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$
Comparing the sets,we see that $A = B^{\prime}$.
Thus,the statement is true.