The book answer is correct
There is the master, mistress and $12$ guests, and in the absence of information to the contrary, it is customary to take chairs to be unnumbered, so there is only one way to seat the hosts opposite to each other.
In effect, the remaining $12$ seats become numbered with respect to the mistress at the $12$ o'clock position, say, and the master at $6$ o'clock position.
In between there are $6$ seats in each half, but the two specified people can only sit together in $2\cdot5$ ways in each half, i.e. $2\cdot10$ ways altogether.
The remaining $10$ guests can then be arranged in $10!$ ways
Putting everything together, ans $= 2*10*10!$
I am not sure I follow your logic $100$%, but here is how I would solve the problem.
You are counting, in a random permutation of $\{1, 2, \dots, n\}$, the number of pairs of positions that have simply swapped places. Call this number $M$. Luckily you only want the expected value $E[M]$ so linearity of expectation comes to the rescue.
For $1 \le i < j \le n$, let $X_{ij}$ be the indicator random variable for whether $i$ and $j$ have swapped places. E.g. if the permutation is $(3,5,1,2,4)$ then $X_{13} = 1$ and all other $X_{ij} = 0$. Then $M = \sum_{1 \le i < j \le n} X_{ij}$.
Meanwhile $\forall i, j: E[X_{ij}] = P(X_{ij} = 1) = {(n-2)! \over n!} = {1 \over n(n-1)}$ because the other $(n-2)$ positions are unconstrained.
So by linearity of expectation (and despite the various $X_{ij}$'s being clearly dependent), we have:
$$E[M] = \sum_{1 \le i < j \le n} E[X_{ij}] = {n \choose 2} \cdot {1 \over n(n-1)} = {1 \over 2}$$
So this is a constant (independent of $n$) as you suspected, but not the value you calculated.
Sanity check: for $n=3$, out of the $6$ permutations, $3$ of them have $M=1$ (choose any of $3$ pairs), and the other $3$ (identity and two rotations) have $M=0$, for an average of $E[M] = 1/2.$
As I said, I am not sure I follow your logic $100$%, but let me try to debug at least one piece of it. In your formula:
$$E[M] = 1 \cdot \frac{{5 \choose 2} 3!}{5!} + 2 \cdot \frac{{5 \choose 2} {3 \choose 2}}{5!} = 1$$
You counted the number of permutations with $1$ pair swapped to be ${5 \choose 2}3!$ but this is incorrect. If the ${5 \choose 2}$ denotes the pair which have swapped, then you need to make sure there is not another swapped pair among the remaining $3$. So you cannot multiply by $3!$ since that allows any permutation among the other $3$. Indeed, among the $3!=6$ permutations, exactly $3$ of them have no swapped pair (identity and the two rotations). So that numerator should have been ${5 \choose 2}\color{red}{3}$ instead.
You counted the number of permutations with $2$ swapped pairs to be ${5\choose 2}{3 \choose 2}$ but this is exactly double counting, since either of the two pairs could have been chosen first. So that numerator should have been ${5 \choose 2}{3 \choose 2}\color{red}{/2}$ instead.
So the correct formula would be:
$$E[M] = 1 \cdot \frac{{5 \choose 2} \color{red}{3}}{5!} + 2 \cdot \frac{{5 \choose 2} {3 \choose 2}\color{red}{/2}}{5!} = {10 \cdot 3 + 2 \cdot 10 \cdot 3 /2 \over 5!} = {60 \over 120} = {1 \over 2}$$
In general, your method finds $P(M=i)$ but I think that is actually a (much?) harder problem than finding $E[M]$ using my method.
Best Answer
Seat the master. It does not matter where since two seating arrangements that differ by a rotation are considered to be equivalent. The mistress must sit opposite the master. Since there are $12$ guests, there are six seats between the master and mistress on each side. If we proceed clockwise around the table from the master, the block of two seats for the two guests who must sit together must either begin in one of the first five seats before we reach the mistress or in one of the first five seats after we reach the mistress. Choose one of those $10$ places for the block of two seats to begin. We have two ways to arrange the specified guests in the block of two seats. The remaining ten guests can be arranged in $10!$ ways in the remaining ten seats as we proceed clockwise around the table from the master. Hence, there are $$2 \cdot 10 \cdot 10!$$ admissible seating arrangements.