jmc

Evaluating the first term, $\frac{12}{7}\cdot \frac{-29}{3}=\frac{-116}{7}$. Since $$-17=\frac{-119}{7}<\frac{-116}{7}<\frac{-112}{7}=-16,$$ the ceiling of $\frac{-116}{7}$ is $-16$.

In the second term, since $$-10=\frac{-30}{3}<\frac{-29}{3}<\frac{-27}{3}=-9,$$ then the floor of $\frac{-29}{3}$ is $-10$. The product of this with $\frac{12}{7}$ is $\frac{-120}{7}$. Since $$-18=\frac{-126}{7}<\frac{-120}{7}<\frac{-119}{7}=-17,$$ the floor of $\frac{-120}{7}$ is $-18$. Hence, the answer is $-16-(-18)=\boxed{2}$.