For the universal set$\{ $$4,5,6,7,8,9,10,11,12,13$$\} $, find its subsets $A,B,C$and $D$
$A$) $\{ $even numbers$\} $
$B$) $\{ $odd numbers greater than $8$$\} $
$C$) Prime numbers
$D$) even numbers less than $10$

Hint: First we have to define what the terms we need to solve the problem are.
A number which is divisible by \[2\]and generates a remainder of \[0\]is called an even number.
Odd numbers are whole numbers that cannot be divided exactly into pairs. Odd numbers, when divided by\[2\], leave a remainder of \[1,3,5,7,9,11,13,15{\text{ }} \ldots \]are sequential odd numbers. Odd numbers have the digits \[1,3,5,7or9\] in their one’s place.
Prime numbers are whole numbers greater than\[1\] , that have only two factors \[1\]and the number itself. Prime numbers are divisible only by the number \[1\]or itself.

Complete step by step answer:
Since we know the definition of even number, odd number and prime number we further approach to find option $A$ which is the set of all even numbers in the given set.
Thus $\{ 4,6,8,10,12\} $are the numbers which are divisible by \[2\]and generates a remainder zero
Hence $A$=$\{ 4,6,8,10,12\} $
Now for option $B$which is the set of all odd numbers also greater than $8$
Thus, odd numbers also greater that $8$ as seen $\{ 9,11,13\} $are the numbers which are cannot divided exactly two pairs and leaves a remainder \[1,3,5, \ldots \]
Hence $B$= $\{ 9,11,13\} $
Now for option $C$ which is a set of all prime number to find from universal set
Since Prime numbers are greater than \[1\], that have only two factors \[1\]and the number itself which are $C$= $\{ 5,7,11,13\} $
And finally, $D$ is the even numbers also less that $10$
which are $D$= $\{ 4,6,8\} $even numbers also less that $10$
Hence $A$=$\{ 4,6,8,10,12\} $, $B$=$\{ 9,11,13\} $, $C$= $\{ 5,7,11,13\} $and $D$= $\{ 4,6,8\} $

Note: We find even numbers, odd numbers, prime numbers all in the universal set only
And $A,B,C$and $D$ are the subsets of the given universal set. Also, the universal set does not contain any repeated elements.