Here, $A=\{1,2\}, B=\{1,2,3,4\}, C=\{5,6\}$ and $D=\{5,6,7,8\}$
(i) $\quad B \cap C=\{1,2,3,4\} \cap\{5,6\}=\phi$
$A \times(B \cap C)=A \times \phi=\phi \quad \ldots (i)$
Now, $A \times B=\{1,2\} \times\{1,2,3,4\}$
$=\{(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4)\}$
$A \times C=\{1,2\} \times\{5,6\}$
$=\{(1,5),(1,6),(2,5),(2,6)\}$
Hence $(A \times B) \cap(A \times C)=\phi \quad \ldots(ii)$
From (i) and (ii), we have
$A \times(B \cap C)=(A \times B) \cap(A \times C)$.
(ii) $A \times C=\{1,2\} \times\{5,6\}$
$=\{(1,5),(1,6),(2,5),(2,6)\}$ $=\{1,2,3,4\} \times\{5,6,7,8\}$
$B \times D=\{1,2,3,4\} \times\{5,6,7,8\}$
$\begin{aligned}=&\{(1,5),(1,6),(1,7),(1,8),(2,5)\\ &(2,6),(2,7),(2,8),(3,5),(3,6) \\ &(3,7),(3,8),(4,5),(4,6),(4,7),(4,8)\} \end{aligned}$
Clearly, $A \times C \subseteq B \times D$
because every element of set $(A \times C)$ is in set $(B \times D)$.