In how many ways can $12$ people be seated at a square table if $90^\circ$ rotations are considered to be equivalent?

Say for instance we are give a table as follows, with the condition that if you were to rotate the table 90 degrees it would still be considered the same.

\begin{bmatrix}&A&B&C&\\L&&&&D\\K&&&&E\\J&&&&F\\&I&H&G\end{bmatrix}

Q1) How many ways can 12 be seated, under this specified condition?

My answer) 3 cases to consider. Someone sits next to the Right of A. Someone sits to the Left of A. A sits between 2 people. Therefore $3*11!$

Q2) How many different seating are possible with A and B must be sitting next to each other?

My answer) A and B must be sitting with each other, however due to 90/180/270 degree turns that eliminates three options. 3*11! - 27*10!

Q3) How many different seating are possible with A and B must NOT be sitting next to each other.

My answer) 3 cases to consider. Someone sits next to the Right of A. Someone sits to the Left of A. A sits between 3 people. We have 9 positions where people can be seated with 3 cases to follow, therefore $9*10! + 9*10! + 9*10! = 27*10!$

Q4) What is the probability that a seating arrangement chosen at random will have A and B not sitting next to each other?

My answer) We know # of people who cannot sit together and total seating arrangements, therefore 27*10! / 3*11!

Ultimately, I'd like to know if logic in Q1 and Q3 is sound, as it would let me answer Q2/Q4 very easily.

How many ways can $12$ people be seated at a square table if $90^\circ$ rotations are considered to be equivalent?

Method 1: When viewed from the opposite side of the table, person A is either in the left, middle, or right seat. There are $11!$ to seat the other eleven people as we proceed clockwise around the table relative to $A$, which yields $3 \cdot 11!$ possible seating arrangements.

Method 2: There are $12!$ ways of seating $12$ people in order. Since $90^\circ$ rotations of seating arrangements are considered to be equivalent, we must divide by $4$, which yields $$\frac{12!}{4} = \frac{12 \cdot 11!}{4} = 3 \cdot 11!$$

How many ways can the $12$ people be seated at a square table if A and B sit in adjacent seats if $90^\circ$ rotations are considered to be equivalent?

We first seat person A. When viewed from the opposite side of the table, person A is either in the left, middle, or right seat. Person B can be seated to the immediate left or immediate right of A. The remaining $10$ people can be seated in $10!$ ways as we move clockwise around the table relative to the block of two seats now occupied by persons A and B. Hence, there are $3 \cdot 2 \cdot 10!$ seating arrangements in which A and B sit in adjacent seats.

How many ways can the $12$ people be seated at a square table if $A$ and $B$ do not sit in adjacent seats if $90^\circ$ rotations are considered to be equivalent?

Subtract the number of ways they can sit in adjacent seats from the total number of seating arrangements.

What is the probability that a seating arrangement chosen at random will have A and B not sitting next to each other if $90^\circ$ rotations are considered to be equivalent?

Divide the preceding result by the total number of seating arrangements, which is equivalent to $1$ minus the probability that they are sitting next to each other.