Recall: $a$ tables, $2b$ people at each table. Label the couples $1,\ldots,ab$. Define the event $C$ as follows:
$$
C = \{\mbox{couples $1,\ldots,k$ are good and all others are bad}\}.
$$
By symmetry, observe that
$$
\mathbb{P}(\mbox{exactly $k$ good couples}) = \left(\begin{array}{c}
ab \\ k
\end{array}\right)\
\mathbb{P}(C).
$$
We'll compute $\mathbb{P}(C)$ with an inclusion-exclusion argument.
Let $\pi_n$ be the probability that couples $1,\ldots,n$ are good (regardless of whether any other couple is good). Note that by symmetry again, $\pi_n$ is the probability that any particular set of $n$ couples are good. Define the events $A$ and $A_n$ as follows for $n=k+1,\ldots,ab$:
\begin{eqnarray}
A &=& \{\mbox{couples $1,\ldots,k$ are good} \} \\
A_n &=& \{\mbox{couples $1,\ldots,k$, and $n$ are good}\}.
\end{eqnarray}
Observe
$$
C = A \setminus \bigcup_{i=k+1}^{ab} A_i.
$$
Inclusion-exclusion:
$$
\mathbb{P}(C) = \mathbb{P}(A) - \sum_{t=1}^{ab-k} \sum_{i_1,\ldots,i_t} (-1)^{t+1} \mathbb{P}(A_{i_1}\cap\cdots\cap A_{i_t}).
$$
Again by symmetry, $\mathbb{P}(A_{i_1}\cap\cdots\cap A_{i_t})=\pi_{k+t}$, and there are $\left(\begin{array}{c}ab - k \\ t\end{array}\right)$ terms of this form in the sum above. Thus, we may write
$$
\mathbb{P}(C) = \sum_{t=0}^{ab-k} (-1)^t\ \left(\begin{array}{c}ab - k \\ t\end{array}\right) \ \pi_{k+t}.
$$
We can move on to computing $\pi_n$.
Let us count the number of ways in which each of the $2ab$ people can be assigned tables in which couples $1,\ldots,n$ are good. Suppose that among these $n$ couples, $n_1$ are seated at table 1, $n_2$ at table 2, $\ldots$, and $n_a$ are seated at table $a$. Thus, $n_1+\cdots +n_a = n$. There are $\left(\begin{array}{c}n \\ n_1,\ldots,n_a\end{array}\right)$ ways of choosing which of the $n$ couples sit at which table. There are $\left(\begin{array}{c}2ab-2n \\ 2b-2 n_1,\ldots,2b-2n_a\end{array}\right)$ ways to choose the tables at which the remaining $2ab-2n$ people sit. Thus, the total number of ways of assigning people to tables in such a way that couples $1,\ldots,n$ are good is
$$
\sum_{n_1+\cdots+n_a=n} \ \left(\begin{array}{c}n \\ n_1,\ldots,n_a\end{array}\right)\ \left(\begin{array}{c}2ab-2n \\ 2b-2 n_1,\ldots,2b-2n_a\end{array}\right).
$$
The total number of ways of assigning all $2ab$ people is just $\frac{(2ab)!}{(2b)!^a}$. Thus,
$$
\pi_n = \frac{(2b)!^a}{(2ab)!} \sum_{n_1+\cdots+n_a=n} \ \left(\begin{array}{c}n \\ n_1,\ldots,n_a\end{array}\right)\ \left(\begin{array}{c}2ab-2n \\ 2b-2 n_1,\ldots,2b-2n_a\end{array}\right).
$$
Alternatively, note that $\pi_n$ can be written
$$
\pi_n = \frac{(2b)!^a \ n! \ (2ab - 2n)!}{(2ab)!} C_n
$$
where $C_n$ is the coefficient of $X^n$ in the polynomial
$$
\left(\sum_{i=0}^b \frac{1}{i! (2b - 2i)!} X^i\right)^a.
$$
Best Answer
The given solution is correct. Let us recall that:
In your case, $X$ is the sum of the $X_i$, where $X_i=1$ if husband and wife of couple $i$ are seated next to each other, and $X_i=0$ otherwise. True, the random variables $X_i$ are not independent, but the only important things are that, for every $i$, $\mathrm E(X_i)=\frac2{25}$, and that there are $13$ couples. Hence indeed $\mathrm E(X)=13\times\frac2{25}$.