There are 5 divisions in the country.
4 divisions sent 3, 4, 5 and 5 representative.
Let representative sent by the 5th division = x
Question: Was the range of the numbers of representatives sent by the five directions greater than 2?
Statement 1: The median of the numbers of representatives sent by the five divisions was greater than the average (arithmetric mean) of these numbers.
If x = 2
Representatives sent = 2, 3, 4, 5, 5.
$$Mean=\frac{2+3+4+5+5}{5}=\frac{19}{5}=3.8$$
Median = 4; hence, median > average
Range = highest rep. - lowest rep. = 5 - 2 = 3; Range > 2.
If x = 5
Representatives sent = 3, 4, 5, 5, 5.
$$Mean=\frac{3+4+5+5+5}{5}=\frac{22}{5}=4.4$$
Median = 5; hence, median > average
Range = highest rep. - lowest rep. = 5 - 3 = 2; Range = 2.
Therefore, the information provided is not enough to ascertain if Range > 2, hence, statement 1 is NOT SUFFICIENT.

Statement 2: The median of the numbers of representatives sent by the five divisions was 4.
For the median of the representative to be equal to 4, 'x' has to be less than or equal to 4, given that the number of representatives is already available (i.e 3, 4, 5, 5).
If x=2, the range > 2 but if x=4, the range = 2.
Therefore, the information provided is not enough to ascertain if the range >2, hence, statement 2 is NOT SUFFICIENT.

Combining both statement together;
Median=4 and median>mean (i.e mean < median)
$$\frac{3+4+5+5+x}{5}<3$$
$$17+x<20$$
$$x<20-17$$
$$x<3$$
If x=2
Range = 5 - 2 = 3
If x=1
Range = 5 - 1 = 4
In both cases, Range >2, hence, both statements combined together are SUFFICIENT.

Answer = C