2. In the plane, there are $h+s$ lines, among which $h$ lines are horizontal, and $s$ lines satisfy:
(1) None of them are horizontal lines;
(2) No two of them are parallel;
(3) Any 3 of these $h+s$ lines do not intersect at the same point.
If these $h+s$ lines exactly divide the plane into 1992 regions, find all pairs of positive integers $(h, s)$.
2. First, calculate the number of regions formed by $s$ lines that are not parallel to each other and no three of which are concurrent.

Let $a_{n}$ represent the number of regions formed by $n$ lines that are not parallel to each other and no three of which are concurrent.
When we add one more line, it intersects the previous $n$ lines at $n$ points, meaning this new line is divided into $n+1$ segments (line segments and rays), and each segment divides the original region into two.
Thus, adding one more line increases the number of regions by $n+1$, so $a_{n+1}=a_{n}+(n+1)$.
It is easy to see that $a_{1}=2$, so $a_{n}=a_{n-1}+n=a_{n-2}+(n-1)+n=\cdots$
$$
=n+(n-1)+\cdots+3+2+2=1+\frac{n(n+1)}{2} .
$$

Therefore, $s$ lines divide the plane into $1+\frac{s(s+1)}{2}$ regions.
Since each horizontal line intersects these $s$ lines at $s$ points, this horizontal line is divided into $s+1$ segments, and each segment divides the original region into two.
That is, adding one horizontal line increases the number of regions by $s+1$.
Thus, after adding $h$ horizontal lines, the number of regions in the plane is $h(s+1)+1+\frac{s(s+1)}{2}=1992$,
which simplifies to $(s+1)(2 h+s)=2 \times 11 \times 181$, so we have

Therefore, the desired pairs of positive integers $(h, s)$ are $(995,1),(176,10),(80,21)$.