9. (10 points) Person A and Person B start from point A at the same time and walk uniformly towards point B, while at the same time Person C starts from point B and walks uniformly towards point A. 20 minutes after departure, A meets C, and immediately turns around. After 10 minutes, A meets B, then A turns around again and heads towards B. As a result, when A reaches B, B has just passed the midpoint of A and B by 105 meters, and C is still 315 meters away from A. The speed of A is $\qquad$ times the speed of B, and the distance between A and B is $\qquad$ meters.

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【Analysis】(1) In 30 minutes, A walked forward 2 portions of the distance and then walked backward 1 portion, totaling 3 portions of the distance, while B only walked 2-1=1 (portion) of the distance; then according to speed $\times$ time $=$ distance, it can be derived that when the time is constant, the ratio of speeds equals the ratio of the distances they travel, to find out how many times A's speed is of B's speed;
(2) Based on A's speed being 3 times B's speed, when B has walked 105 meters past the midpoint of $A B$, if A had not turned back, A should have traveled: 3 half-distances + 315 meters; that is, A would have traveled 1 full distance plus 1 half-distance + 315 meters. In the 20 minutes from when A meets C and turns back to meet B and then returns to the meeting point of A and C, A can walk 1 half-distance + 315 meters; so when A meets C, A has already traveled 1 half-distance + 315 meters, and C has traveled 1 half-distance - 315 meters; later, if A does not stop, after traveling 3 half-distances + 315 meters, C has traveled 2 half-distances - 315 meters; after A and C meet, A travels 2 half-distances, and C travels 1 half-distance; so A's speed is 2 times C's speed; that is, when A and C meet, A has traveled $\frac{2}{2+1}=\frac{2}{3}$ of the total distance, which is 315 meters more than 1 half-distance, according to this, the distance between $A$ and $B$ can be found.

【Solution】Solution: (1) A's speed is B's speed:
$$
(2+1) \div(2-1)=3 \text { (times); }
$$
(2) When B has walked 105 meters past the midpoint of $A B$, the distance A has traveled is more than 1 full distance by 1 half-distance + 315 meters. When A and C meet, A has traveled $\frac{2}{2+1}=\frac{2}{3}$ of the total distance,
so the distance between $A$ and $B$ is:
$$
\begin{array}{l}
315 \div\left(\frac{2}{3}-\frac{1}{2}\right) \\
=315 \div \frac{1}{6} \\
=1890 \text { (meters) }
\end{array}
$$

Answer: A's speed is 3 times B's speed, the distance between $A$ and $B$ is 1890 meters.
The answer is: 3, 1890.