A cylinder's mass (m) varies jointly as the square of its radius (r) and its height (h). If the radius is tripled and the height is halved, what is the percentage change in mass?

Let the initial mass be $m_1$, initial radius $r_1$, and initial height $h_1$.

$$m_1 = kr_1^2h_1$$

The new radius $r_2 = 3r_1$.
The new height $h_2 = \frac{1}{2}h_1$.
The new mass $m_2$ is:

$$m_2 = k(3r_1)^2\left(\frac{1}{2}h_1\right)$$ $$m_2 = k(9r_1^2)\left(\frac{1}{2}h_1\right)$$ $$m_2 = \frac{9}{2}kr_1^2h_1$$

Substitute $m_1 = kr_1^2h_1$ into the equation for $m_2$:

$$m_2 = \frac{9}{2}m_1$$

$$\text{Percentage Change} = \frac{m_2 - m_1}{m_1} \times 100\%$$ $$\text{Percentage Change} = \frac{\frac{9}{2}m_1 - m_1}{m_1} \times 100\%$$ $$\text{Percentage Change} = \left(\frac{9}{2} - 1\right) \times 100\%$$ $$\text{Percentage Change} = \left(\frac{9}{2} - \frac{2}{2}\right) \times 100\%$$ $$\text{Percentage Change} = \frac{7}{2} \times 100\%$$ $$\text{Percentage Change} = 3.5 \times 100\%$$ $$\text{Percentage Change} = \boxed{350\%}$$

Substitute $m = 990$, $r = 3$, and $h = 7$ into the formula $m = kr^2h$:

$$990 = k(3)^2(7)$$ $$990 = k(9)(7)$$ $$990 = 63k$$

$$k = \frac{990}{63}$$

Divide both numerator and denominator by 9:

$$k = \frac{110}{7}$$

Substitute the value of $k$ back into the general equation:

$$m = \boxed{\frac{110}{7}r^2h}$$

Substitute $r = 3.5$ and $h = 5$:

$$m = \frac{110}{7}(3.5)^2(5)$$ $$m = \frac{110}{7}\left(\frac{7}{2}\right)^2(5)$$ $$m = \frac{110}{7}\left(\frac{49}{4}\right)(5)$$

$$m = \frac{110 \times 49 \times 5}{7 \times 4}$$

Cancel out 7 from 49:

$$m = \frac{110 \times 7 \times 5}{4}$$ $$m = \frac{110 \times 35}{4}$$

Divide 110 by 2 and 4 by 2:

$$m = \frac{55 \times 35}{2}$$ $$m = \frac{1925}{2}$$ $$m = \boxed{962.5\text{ g}}$$