Determine the radius of the can.

The diameter of a can is $6 \text{ cm}$.

$$r = \frac{\text{Diameter}}{2} = \frac{6 \text{ cm}}{2} = 3 \text{ cm}$$

Given $\pi = 3.142$ and height ($h$) = $8.84 \text{ cm}$.

$$V = \pi r^2 h$$ $$V = 3.142 \times (3 \text{ cm})^2 \times 8.84 \text{ cm}$$ $$V = 3.142 \times 9 \text{ cm}^2 \times 8.84 \text{ cm}$$ $$V = 28.278 \text{ cm}^2 \times 8.84 \text{ cm}$$ $$V = 249.90792 \text{ cm}^3$$

$$V \approx 250 \text{ cm}^3$$

The volume of one can of cold drink is $\boxed{\text{250 cm}^3}$.

From 9.1, the volume of one can is $249.90792 \text{ cm}^3$.

$$\text{Capacity} = 249.90792 \text{ ml}$$

$$\text{Capacity} \approx 250 \text{ ml}$$

The capacity of one can of cold drink is $\boxed{\text{250 ml}}$.

The problem states that "Two six-pack cans fit exactly on top of each other in the cooler bag."
The height of one can is $8.84 \text{ cm}$.

$$\text{Height}_{\text{cooler bag}} = 2 \times \text{Height}_{\text{can}}$$ $$\text{Height}_{\text{cooler bag}} = 2 \times 8.84 \text{ cm}$$ $$\text{Height}_{\text{cooler bag}} = 17.68 \text{ cm}$$

$$\text{Height}_{\text{cooler bag}} \approx 18 \text{ cm}$$

The height of the cooler bag is $\boxed{\text{18 cm}}$.

$$\text{Height}_{\text{cooler bag}} = 2 \times 8.84 \text{ cm}$$ $$\text{Height}_{\text{cooler bag}} = 17.68 \text{ cm}$$

$$\text{Height}_{\text{cooler bag}} \approx 18 \text{ cm}$$

The height of the cooler bag is $\boxed{\text{18 cm}}$.