Arrangements around a square, People not sit across diagonally - Extension The first two parts of this question are exact same as the Counting the arrangements of 8 people around a square table? and have already been answered but my struggle is in the final part of my problem which is an extension of the problem cited.
Basically, the final part of the question ask how to arrange 8 people around a square with 2 people on each side such that 2 people who hate each other won't sit on the same side or on parallel side, i.e, they cannot sit across directly or diagonally.
I solved the first two parts, where part a my answer was $$7!\times2$$ and for the second part my answer was $$2\times(7!-6!)$$ the second answer might be wrong but I am confident about the answer to the first part
 A: 
In how many ways can eight people be seated at a square table if there are two people per side up to rotations by $90^\circ$, $180^\circ$, $270^\circ$, or $360^\circ$?

Your answer is correct.
Ann can be seated in eight ways.  As we proceed clockwise around the table from Ann, the others can be seated in $7!$ ways.  Dividing by four to account for equivalent rotations yields
$$\frac{8 \cdot 7!}{4} = 2 \cdot 7!$$
as you found.

In how many ways can eight people be seated at a square table if Ann and Bob do not sit on the same side?

Again, your answer is correct.
Ann can be seated in eight ways.  Since Bob does not sit on the same side of the table, he may be seated in six ways.  The remaining six people may fill the remaining six seats in $6!$ ways as we proceed clockwise around the table from Ann.  Dividing by four to account for equivalent rotations yields
$$\frac{8 \cdot 6 \cdot 6!}{4} = 2 \cdot 6 \cdot 6!$$ 
which agrees with the answer you obtained by subtracting those arrangements in which Ann and Bob sit on the same side of the table from the total number of arrangements.

In how many ways can eight people be seated at a square table if Ann and Bob do not sit on the same side of the table or across from each other?

If we interpret this to mean that Bob cannot sit on the same side as Ann or directly across from Ann, then she can be seated in eight ways, he can be seated in five ways, and the others may be seated in $6!$ ways as we proceed clockwise around the table from Ann.  Dividing by $4$ to account for rotational equivalence yields
$$\frac{8 \cdot 5 \cdot 6!}{4} = 2 \cdot 5 \cdot 6!$$
If we instead interpret the question to mean that Bob cannot sit on the same side as Ann or the side opposite to Ann, then there are only four ways to seat Bob, so we would have 
$$\frac{8 \cdot 4 \cdot 6!}{4} = 8 \cdot 6!$$
admissible seating arrangements.  
Of course, among these arrangements are those in which the two people who hate each other share a corner of the table, but the author of the problem has not excluded that possibility.
