Let A = $\left\{ {1,2} \right\}$ , B = $\left\{ {1,2,3,4} \right\}$ , C = $\left\{ {5,6} \right\}$ and D = $\left\{ {5,6,7,8} \right\}$ . Verify that
(i) ${\text{A}} \times \left( {{\text{B}} \cap {\text{C}}} \right) = \left( {{\text{A}} \times {\text{B}}} \right) \cap \left( {{\text{A}} \times {\text{C}}} \right)$
(ii) ${\text{A}} \times {\text{C }}$ is a subset of ${\text{B}} \times {\text{D}}$

Hint: As in this question we have to verify the equation as ${\text{A}} \times \left( {{\text{B}} \cap {\text{C}}} \right) = \left( {{\text{A}} \times {\text{B}}} \right) \cap \left( {{\text{A}} \times {\text{C}}} \right)$ First find the value of ${\text{A}} \times \left( {{\text{B}} \cap {\text{C}}} \right)$ from the given values as A = $\left\{ {1,2} \right\}$ , B = $\left\{ {1,2,3,4} \right\}$ , C = $\left\{ {5,6} \right\}$ and D = $\left\{ {5,6,7,8} \right\}$ Now find the value of $\left( {{\text{A}} \times {\text{B}}} \right) \cap \left( {{\text{A}} \times {\text{C}}} \right)$ equate it if it is equal then it is verify ,
In part (ii) If ${\text{A}} \times {\text{C }}$ is a subset of ${\text{B}} \times {\text{D}}$ then in set ${\text{A}} \times {\text{C }}$have all the element in ${\text{B}} \times {\text{D}}$.

Complete step-by-step answer:
(i)
In this we have to verify the ${\text{A}} \times \left( {{\text{B}} \cap {\text{C}}} \right) = \left( {{\text{A}} \times {\text{B}}} \right) \cap \left( {{\text{A}} \times {\text{C}}} \right)$ where A = $\left\{ {1,2} \right\}$ , B = $\left\{ {1,2,3,4} \right\}$ , C = $\left\{ {5,6} \right\}$ and D = $\left\{ {5,6,7,8} \right\}$ is given ,
${\text{B}} \cap {\text{C}}$ mean that the common digit between B = $\left\{ {1,2,3,4} \right\}$ and C = $\left\{ {5,6} \right\}$
So in ${\text{B}} \cap {\text{C}}$ there is no common values hence ,
${\text{B}} \cap {\text{C}}$= $\phi $
And ${\text{A}} \times \left( {{\text{B}} \cap {\text{C}}} \right) = {\text{A}} \times \phi $= $\phi $ hence it is an empty set .
${\text{A}} \times \left( {{\text{B}} \cap {\text{C}}} \right) = \phi $ ..........(i)

Now for the $\left( {{\text{A}} \times {\text{B}}} \right) \cap \left( {{\text{A}} \times {\text{C}}} \right)$
Hence the value of ${\text{A}} \times {\text{B}}$ = $\left\{ {(1,1),(1,2)(1,3),(1,4),(2,1),(2,2),(2,3),(2,4)} \right\}$
and the value of ${\text{A}} \times {\text{C}}$ = $\left\{ {(1,5),(1,6),(2,5),(2,6)} \right\}$
$\left( {{\text{A}} \times {\text{B}}} \right) \cap \left( {{\text{A}} \times {\text{C}}} \right)$ is that relation which are common in ${\text{A}} \times {\text{B}}$ and ${\text{A}} \times {\text{C}}$ ,
So there is no common relation among them hence ,
$\left( {{\text{A}} \times {\text{B}}} \right) \cap \left( {{\text{A}} \times {\text{C}}} \right)$ = $\phi $ ........(ii)
Hence from (i) and (ii) we say that ${\text{A}} \times \left( {{\text{B}} \cap {\text{C}}} \right) = \left( {{\text{A}} \times {\text{B}}} \right) \cap \left( {{\text{A}} \times {\text{C}}} \right)$

Now for the part (ii) we have to verify that the ${\text{A}} \times {\text{C }}$ is a subset of ${\text{B}} \times {\text{D}}$
If ${\text{A}} \times {\text{C }}$ is a subset of ${\text{B}} \times {\text{D}}$ then in set ${\text{A}} \times {\text{C }}$have all the element in ${\text{B}} \times {\text{D}}$so for this
${\text{A}} \times {\text{C }}$ = $\left\{ {(1,5),(1,6),(2,5),(2,6)} \right\}$ and
${\text{B}} \times {\text{D}}$ = $\left\{ \
  (1,5),(1,6),(1,7),(1,8), \\
  (2,5),(2,6),(2,7),(2,8), \\
  (3,5),(3,6),(3,7),(3,8), \\
  (4,5),(4,6),(4,7),(4,8), \\
\ \right\}$
 From above we can find that the all element of ${\text{A}} \times {\text{C }}$ is present in ${\text{B}} \times {\text{D}}$ hence it is proof that ${\text{A}} \times {\text{C }}$ is a subset of ${\text{B}} \times {\text{D}}$

Note: Empty Relation
An empty relation (or void relation) is one in which there is no relation between any elements of a set as in the question ${\text{A}} \times \left( {{\text{B}} \cap {\text{C}}} \right) = \phi $ hence it is a empty Relation or void relation .