Answer:
For the first group we have the pairs:
year population
0 20
1 80
2 320
3 1,280
4 5,120
5 20,480
Here we can see that the population quadruples each year
(4*20 = 80, 80*4 = 320, 320*4 = 1,280, etc...)
then the population equation is:
P(0) = 20
P(1) = 20*4
P(2) = (20*4)*4 = 20*4^2
We already can see the pattern, then we can write this relationship as:
P(t) = A*(4)^(t)
Where:
t represents time in years, and A is the initial population, that we know it is 20, then:
P(t) = 20*(4)^t
This is the function that represents the table.
B) Now we have a group of 20 squirrels and the population triples each year, with the same reasoning than before we can write the equation that models this situation as:
Q(t) = 20*(3)^t
C) Now, if the initial population of the second group is 40, the equation becomes:
Q(t) = 40*(3)^t
The population by year 3 is given by replacing t by 3, then:
Q(3) = 40*(3)^3 = 1080
And the population of the other group in year 3 is seen in the table, it is 1,280, then the population of the first group is bigger by year 3, and it is greater by:
1,280 - 1,080 = 200
So the first group is larger by 200 squirrels.
÷ 
Use the recriprocal of 
Multiply (3 x 8) and (4 x 1)
Divide