A Graph logical task - Ramsey theory I have already asked in puzzling.SE this task but from there they pointed to me that it is more a graph question than a puzzle:
https://puzzling.stackexchange.com/questions/98396/a-perfect-world-completely-friends-or-completely-not-friends
The task is the following:

In a perfect world two people are either friends or not friends. In this perfect world the population is exactly 6 people.
In this perfect world a group of $n$ people is called completely friends when each person in the group is friends with every other person in the group.
In this perfect world a group of $n$ people is called completely not friends when each person in the group is not friends with every other person in the group.
Which of the below statements are always true?

*

*We can always find a group of 3 which is either completely friends or completely not friends.


*If we have a group of 4 which are completely friends then the number of friendships is higher than the number of not friendships. (A friendship is when two people are friends. A not friendship is when two people are not friends.)


*If we have a group of 3 which are completely not friends, there is no way for each person to be friend with exactly two others.


*If we know that each two people have at least one common friend, then this means that we have a group of 4 which are completely friends.


*There is no way for each two people to have exactly one common friend (from the other four).


*There is no way for each two people to have exactly one common not friend (from the other four).

 A: If we want to solve your question "mathematically" we need to rephrase your problem in standard math terms, in our case graph theory terms. 
We are building a graph on 6 vertices (the total world propulation, each vertex is one person), and we put an edge between two vertices if and only if they are friends.
For example in a world where everybody is friend with everybody, we would have what we call the complete graph on 6 vertices, $K_6$ :

Question 2 : If we have a group of 4 which are completely friends then the number of friendships is higher than the number of not friendships
False. Here as the statement is a positive conjecture, a counterexample is sufficient. The extreme case of $K_4$ and two isolated vertices works : We have 4 persons all friends one with another, and two people knowing nobody.

In this case there are 6 edges, over a possible total number of edges of 15 (the case $K_6$ has 15 edges). Therefore there are 9 "non-edges", more than edges.
Question 4 If we know that each two people have at least one common friend, then this means that we have a group of 4 which are completely friends.
False Here is a counterexample

The idea is derived from the frienship graphs as seen on the next question.
Question 5 : There is no way for each two people to have exactly one common friend (from the other four).
True. This come from the friendship theorem of Paul Erdős, Alfréd Rényi, and Vera T. Sós (1966) - links to proofs are given here at wiki. It states that if a finite graph has the property that every two vertices have exactly one neighbour in common then there must be a vertex adjacent to everybody, and the graph is one of the so-called friendship graphs. 
In your case it means that we would need to have someone being friend with everybody. But then in order to comply with the hypothesis, we need to pair the other vertices two-by-two, hence we need the total number of vertices to be odd. Which is not the case here as it is 6. Therefore it is not possible. (or said otherwise the Friendship graphs have odd order).
Question 6 : There is no way for each two people to have exactly one common not friend (from the other four).
True. This question is exactly the same question than the 5th one. Indeed we defining your graph, instead of putting an edge when two people are friend, put an edge when two people are not friends. Then question 6 asks for the same condition than our previous question 5 : a graph on 6 vertices where any two vertices have exactly one common neighbour, it is impossible as we have seen.
