1. In still water, the speed of boat A is twice the speed of boat B. Boats A and B start from points $A$ and $B$ respectively at the same time, moving towards each other, and meet at a point where the distances from $A$ and $B$ are in the ratio of 3:1. If boats A and B start from points $B$ and $A$ respectively at the same time, moving towards each other, the ratio of the distances from $A$ and $B$ when they meet is $\qquad$.
【Analysis】Given that the speed of boat A is twice that of boat B, let's assume the speed of boat B in still water is $x$, then the speed of boat A is $2x$, and the water speed is $y$. According to the problem, when boats A and B meet while traveling towards each other, the ratio of the distances from points $A$ and $B$ is 3:1. This means traveling from $A$ to $B$ is downstream, and from $B$ to $A$ is upstream. Therefore, the speed ratio at the first meeting is: $(2x+y):(x-y)=3:1$, from which we can derive $x=4y$; If boats A and B start from points $B$ and $A$ respectively and travel towards each other, the speed ratio at the second meeting is: $(2x-y):(x+y)$. Using $x=4y$, we can find the ratio of the distances from points $A$ and $B$ at the meeting point.

【Solution】Let the speed of boat B in still water be $x$, then the speed of boat A is $2x$, and the water speed is $y$. The speed ratio at the first meeting: $(2x+y):(x-y)=3:1$, from which we can derive $x=4y$;

The speed ratio at the second meeting: $(2x-y):(x+y)$,
Since $x=4y$,
Then $(2x-y):(x+y)=(2 \times 4y):(4y+y)=7:5$, which means the ratio of the distances from points $A$ and $B$ at the meeting point is $5:7$.

Therefore, the answer is: 5:7.