【Solution】Solution: If Jia takes the next number as 1, then Yi takes 4, in the next round, no matter what Jia takes, Jia cannot form an arithmetic sequence, and in the next round, whatever number Yi takes can form an arithmetic sequence $(4, 5, 6$ or $3, 4, 5)$, Jia loses; if Jia takes the next number as 2, then Yi takes 4, similarly, Jia loses; if Jia takes the next number as 3, then Yi takes 6, similarly, Jia loses;

If Jia takes the next number as 4, then Yi takes 6 (otherwise, Jia can win directly by taking 6 in the next round), then Jia can only take 7 (otherwise, Yi can win directly by taking 7 in the next round), in this way, Yi cannot form an arithmetic sequence in this round, in the next round, taking 1 or 9 can form an arithmetic sequence (1, 4, 7 or 7, 8, 9), Jia wins;

If Jia takes the next number as 6, then no matter what Yi takes, Jia can form an arithmetic sequence by taking another number $(4, 6, 8$ or $6, 7, 8)$;

If Jia takes the next number as 7, then no matter what Yi takes, Jia can form an arithmetic sequence by taking another number (6, 7, 8 or $7, 8, 9)$;

If Jia takes the next number as 9, then Yi takes 7 (otherwise, Jia can win directly by taking 7 in the next round), then in the next round, no matter what Jia takes, Yi can form an arithmetic sequence by taking another number (3, 5, 7 or $5, 6, 7$.

In summary, to ensure Jia's certain victory, the next number Jia can take is 4, 6, 7, the product of all possible values is:
$$
4 \times 6 \times 7=168 \text {. }
$$

Therefore, the answer is: 168.
11. (12 points) Two people, A and B, take turns choosing different numbers from the natural numbers $1 \sim 9$. Numbers that have been taken by the opponent cannot be taken again. The first person to have three numbers that form an arithmetic sequence wins. A starts by taking 8, and B then takes 5. To ensure A's certain victory, the product of all possible values that A can take next is $\qquad$ .