What best can be concluded about the number of players coached by Zara?

We are told that each coach had at least two players. Clue 1 says that Xena trained more people than Yuki. 
It must be the case that Yuki trained only two people. If Yuki had trained three, then Xena would have trained at least four players, leaving only one for Zara. 

Hence, Yuki trained two people. 

Coming to the scores given to the players themselves:

We are given that only 5 and 7 received a sam rating; everyone else received a distinct rating. 
We are also given their average to be 4 (clue 4), giving the sum of all the scores they got to be 32
The sum of all numbers from 1 to 7 would be $$\frac{8}{2}\times\ 7=28$$, hence the repeated score must be 32-28 = 4

Thus, the score of 5 and 7 was 4

Clue 5 says that player 2 got the highest score, which is 7

Clue 7 says that player 4 got a score double that of player 8 but less than player 5; the only possibility is player 4 getting 2 and player 8 getting 1. 

Now, considering clues 2, 3 and 6:

We are given the same coach trained in 5 and 7. And 2, 3, and 5 were trained by different coaches, with 1 and 4 being trained by the same coach. 
Cheer 6 informs us about the average number of players the coaches train. 

We know that Yuki trained only two players. 
Let's take the average score of Yuki's players to be x

The average of Xena's players would be x/2, and that of Zara's players would be x-2

The total score of Yuki's players would be 2x. 
The total score of Xena's players would be 3x/2 or 2x. 
The total score of Zara's players would be 3x-6 or 2x-4

We have two cases to consider:

Let's say Xena and Zara had three players each. 
The total score of Yuki's players would be 2x.
The total score of Xena's players would be 3x/2.
The total score of Zara's players would be 3x-6. 

The sum of all these scores would be $$\frac{13x}{2}-6$$ which should be equal to 32
This would give us the value of x as $$\frac{13x}{2}=38$$
Which would give a non-integral value of 2s, that is, the sum of Yuki's player's score. 

Hence, this must not be the case. 

The other possibility:

The total score of Yuki's players would be 2x.
The total score of Xena's players would be 2x.
The total score of Zara's players would be 2x-4.

The sum of all these scores would be 6x-4, which should be equal to 32
This would give the value of x as 6

Hence, the sum of all of Yuki's players would be 12
The sum of all of Xena's players would be 12
The sum of all of Zara's players would be 8

Yuki has only two players whose scores add up to 12; the only combination possible is scores 7+5, where seven were scored by player 2. Hence, player two must be under Yuki. 

Zara got a total of 8 scores, with 7 and 5 gone. The combinations that could get this score are 2+6 and 4+4
The score of 2 is obtained by player 4, which must come with player 1
It is possible that 1 could have gotten a score 6
But then we run into a contradiction: players 3 and 5 would end up under the same coach. 

Hence, Zara must have gotten 8 through 4+4 with players 5 and 7. 

All the remaining scores must be with Xena, adding up to 12, which is the case. 

4 and 1 must be present together, and 3 must be present in Xena as well, giving us the arrangement. 

We can see that Zara had two players. 

Therefore, Option B is the correct answer.