This looks like an error in the book: your answer appears to be correct.
We can count the allowed seatings directly. Suppose that Henry sits next to Tanya. Then there is one seat forbidden to Wilson and Nancy, so there are $\binom72$ ways to choose seats for Wilson and Nancy. There are $2$ ways to seat Wilson and Nancy in those $2$ seats and $6!$ ways to seat the unnamed people. Finally, Henry can be on either side of Tanya, so there are altogether $2\cdot2\cdot6!\cdot\binom72$ acceptable seatings with Henry next to Tanya.
Now suppose that Henry is not seated next to Tanya. Then there are $2$ seats forbidden to Wilson and Nancy, so there are $\binom62$ ways to choose their seats. As before there are $2$ ways to seat them in those two seats and $6!$ ways to seat the unnamed people. Finally, there are $7$ possible choices for Henry’s seat, so there are altogether $7\cdot2\cdot6!\cdot\binom62$ acceptable seatings with Henry not next to Tanya. Altogether, then, there are