The set of ages of Anju's family members is A = 6, 11, 36, 40. What is the cardinal number of the set A? Make a subset by taking any two elements of set A. Is the set A finite or infinite set? Give reason.

The elements are 6, 11, 36, 40. There are 4 distinct elements.  
The cardinal number of set A is $\boxed{4}$.

Set A has 4 elements, which is a countable number.  
Set A is a **finite set** because it has a definite and countable number of elements.

Natural numbers are positive integers starting from 1. So, numbers less than 5 are 1, 2, 3, 4.  
The elements of set B are $\boxed{{1, 2, 3, 4}}$.

A = {2, 4, 6, 8} and B = {1, 2, 3, 4}. The elements are not exactly the same, so they are not equal sets.

Cardinal number of A, $n(A) = 4$.  
Cardinal number of B, $n(B) = 4$.  
Since $n(A) = n(B)$, sets A and B are **equivalent sets**.

A = {2, 4, 6, 8} and B = {1, 2, 3, 4}.  
The common elements are 2 and 4.  
Since they have common elements, sets A and B are **overlapping sets**.

The numbers that divide 6 evenly are 1, 2, 3, and 6.  
The elements of set M are $\boxed{{1, 2, 3, 6}}$.

L = {2, 3, 5, 7} has 4 elements (finite).  
M = {1, 2, 3, 6} has 4 elements (finite).  
N = {1, 2, 3, ...} indicates that the numbers continue indefinitely (uncountable).  
Set **N** is an **infinite set**.

L = {2, 3, 5, 7} and M = {1, 2, 3, 6}. The elements are not exactly the same, so they are not equal sets.

Cardinal number of L, $n(L) = 4$.  
Cardinal number of M, $n(M) = 4$.  
Since $n(L) = n(M)$, sets L and M are **equivalent sets**.

Square numbers are $1^2=1, 2^2=4, 3^2=9, \dots$.  
For $1 \le x < 5$, the square numbers are 1 and 4.  
So, S = {1, 4}.

$n(S) = 2$.

Number of subsets $= 2^2 = 4$.  
There are $\boxed{4}$ possible subsets of S.

The subsets are $\boxed{\emptyset, \{1\}, \{4\}, \{1, 4\}}$.