A dustpan has a triangular cross-section with base 4 cm and height 3 cm. Its length is 7 cm. Calculate the inner surface area to be painted.

The base of the triangle is $4 \text{ cm}$ and the height is $3 \text{ cm}$.

$$A_{\text{triangles}} = 2 \times \frac{1}{2} \times \text{base} \times \text{height} = 2 \times \frac{1}{2} \times 4 \text{ cm} \times 3 \text{ cm} = 12 \text{ cm}^2$$

The dimensions are $4 \text{ cm}$ by $7 \text{ cm}$.

$$A_{\text{bottom}} = \text{length} \times \text{width} = 4 \text{ cm} \times 7 \text{ cm} = 28 \text{ cm}^2$$

The dimensions are $3 \text{ cm}$ by $7 \text{ cm}$.

$$A_{\text{vertical}} = \text{height} \times \text{length} = 3 \text{ cm} \times 7 \text{ cm} = 21 \text{ cm}^2$$

The dimensions are $5 \text{ cm}$ (hypotenuse of the triangle) by $7 \text{ cm}$.

$$A_{\text{slanted}} = \text{hypotenuse} \times \text{length} = 5 \text{ cm} \times 7 \text{ cm} = 35 \text{ cm}^2$$

$$A_{\text{total}} = A_{\text{triangles}} + A_{\text{bottom}} + A_{\text{vertical}} + A_{\text{slanted}}$$ $$A_{\text{total}} = 12 \text{ cm}^2 + 28 \text{ cm}^2 + 21 \text{ cm}^2 + 35 \text{ cm}^2 = 96 \text{ cm}^2$$

The surface area of the dust pan that was painted is $\boxed{\text{96 cm}^2}$.

:
To calculate the external surface area of the wedge, we need to find the area of all its faces: two triangular faces, the bottom rectangular face, the vertical rectangular face, and the slanted rectangular face.

The base of the triangle is $12 \text{ cm}$ and the height is $5 \text{ cm}$.

$$A_{\text{triangles}} = 2 \times \frac{1}{2} \times \text{base} \times \text{height} = 2 \times \frac{1}{2} \times 12 \text{ cm} \times 5 \text{ cm} = 60 \text{ cm}^2$$

The dimensions are $12 \text{ cm}$ by $20 \text{ cm}$.

$$A_{\text{bottom}} = \text{length} \times \text{width} = 12 \text{ cm} \times 20 \text{ cm} = 240 \text{ cm}^2$$

The dimensions are $5 \text{ cm}$ by $20 \text{ cm}$.

$$A_{\text{vertical}} = \text{height} \times \text{length} = 5 \text{ cm} \times 20 \text{ cm} = 100 \text{ cm}^2$$

The dimensions are $13 \text{ cm}$ (hypotenuse of the triangle) by $20 \text{ cm}$.

$$A_{\text{slanted}} = \text{hypotenuse} \times \text{length} = 13 \text{ cm} \times 20 \text{ cm} = 260 \text{ cm}^2$$

$$A_{\text{total}} = A_{\text{triangles}} + A_{\text{bottom}} + A_{\text{vertical}} + A_{\text{slanted}}$$ $$A_{\text{total}} = 60 \text{ cm}^2 + 240 \text{ cm}^2 + 100 \text{ cm}^2 + 260 \text{ cm}^2 = 660 \text{ cm}^2$$

The external surface area of the wedge is $\boxed{\text{660 cm}^2}$.

:
To calculate the surface area of the canvas used for the tent, we need to find the area of the two triangular faces and the two slanted rectangular faces. The base of the tent (on the ground) is not made of canvas.

The base of the triangle is $1.8 \text{ m}$, so half the base is $0.9 \text{ m}$. The slanting edge is $1.5 \text{ m}$. Using the Pythagorean theorem:

$$\text{height}^2 + (0.9 \text{ m})^2 = (1.5 \text{ m})^2$$ $$\text{height}^2 + 0.81 \text{ m}^2 = 2.25 \text{ m}^2$$ $$\text{height}^2 = 2.25 \text{ m}^2 - 0.81 \text{ m}^2 = 1.44 \text{ m}^2$$ $$\text{height} = \sqrt{1.44 \text{ m}^2} = 1.2 \text{ m}$$

The dimensions of each slanted face are the slanting edge ($1.5 \text{ m}$) by the length of the tent ($3 \text{ m}$).

$$A_{\text{slanted faces}} = 2 \times (\text{slanting edge} \times \text{length})$$ $$A_{\text{slanted faces}} = 2 \times (1.5 \text{ m} \times 3 \text{ m}) = 2 \times 4.5 \text{ m}^2 = 9 \text{ m}^2$$

$$A_{\text{total canvas}} = A_{\text{triangles}} + A_{\text{slanted faces}}$$ $$A_{\text{total canvas}} = 2.16 \text{ m}^2 + 9 \text{ m}^2 = 11.16 \text{ m}^2$$

The surface area of the canvas used is $\boxed{\text{11.16 m}^2}$.