1. For negative numbers $a, b$ and $c$, the equalities $\frac{c}{a+b}=2$ and $\frac{c}{b-a}=3$ hold. Which is greater: $c$ or $a$?
Answer: $c<a$.

Solution. From the given equalities, it follows that:

$$
2(a+b)=c=3(b-a) \Rightarrow b=5a \Rightarrow c=12a
$$

Since the number $a$ is negative, multiplying it by 12 makes it smaller. Therefore, $c<a$.

Comment. In the absence of a complete solution, if all transformations are correctly performed and the two unknowns are expressed in terms of the third, but the final answer is incorrect due to the negativity of the numbers not being taken into account - 2 points. If the two unknowns were expressed in terms of the third (even if incorrectly), and the negativity of the numbers was taken into account in the final answer - 3 points.