The base of a trapezoid is $a$, the parallel side is $b$, and the height is $m$. Divide the height into three equal parts $s$ and draw lines parallel to the base through the division points. What are the areas of the trapezoids thus formed?
Let $x$ and $y$ be the two parallel lines drawn through the division points. Then

$$
a-x=y-b=\frac{a-b}{3}
$$

or

$$
x=\frac{2 a+b}{3}
$$

and

$$
y=\frac{a+2 b}{3}
$$

Thus, the areas of the resulting trapezoids are:

$$
\begin{aligned}
t_{1} & =\frac{a+x}{2} \cdot \frac{m}{3}=\frac{5 a+b}{3} \cdot \frac{m}{6} \\
t_{2} & =\frac{x+y}{2} \cdot \frac{m}{3}=(a+b) \frac{m}{6}
\end{aligned}
$$

and

$$
t_{3}=\frac{y+b}{2} \cdot \frac{m}{3}=\frac{a+5 b}{3} \cdot \frac{m}{6}
$$

Number of solutions: 28.