A square-based pyramid has a volume of 384 m3 and a base side length of 12 m. Calculate (a) the number of triangular faces, (b) the height of the pyramid, and (c) the slant height.

The length of each side of the base is $l = 12 \text{ m}$.
The base area $A_b$ is given by:

$$A_b = l^2 = (12 \text{ m})^2 = 144 \text{ m}^2$$

The volume $V$ of a pyramid is given by $V = \frac{1}{3} A_b h$.
Given $V = 384 \text{ m}^3$:

$$384 \text{ m}^3 = \frac{1}{3} (144 \text{ m}^2) h$$ $$384 \text{ m}^3 = 48 \text{ m}^2 \cdot h$$

$$h = \frac{384 \text{ m}^3}{48 \text{ m}^2} = 8 \text{ m}$$

The height of the tent is 8 m.

$$\boxed{8 \text{ m}}$$

The height $h = 8 \text{ m}$ (from part b) and half the base side length is $\frac{l}{2} = \frac{12 \text{ m}}{2} = 6 \text{ m}$.
Using the Pythagorean theorem, $s^2 = h^2 + \left(\frac{l}{2}\right)^2$:

$$s^2 = (8 \text{ m})^2 + (6 \text{ m})^2$$ $$s^2 = 64 \text{ m}^2 + 36 \text{ m}^2$$ $$s^2 = 100 \text{ m}^2$$ $$s = \sqrt{100 \text{ m}^2} = 10 \text{ m}$$

The base of each triangle is $l = 12 \text{ m}$ and the slant height is $s = 10 \text{ m}$.
Area of one triangle $= \frac{1}{2} \times \text{base} \times \text{slant height}$

$$A_{\text{triangle}} = \frac{1}{2} \times 12 \text{ m} \times 10 \text{ m} = 60 \text{ m}^2$$

Since there are 4 triangular faces, the total cloth required is:

$$\text{Total cloth} = 4 \times A_{\text{triangle}} = 4 \times 60 \text{ m}^2 = 240 \text{ m}^2$$

The amount of cloth required to construct the tent is 240 m$^2$.

$$\boxed{240 \text{ m}^2}$$