A) Find the slant height of the cone correct to 2 decimal places. b) Find the total surface area of the solid Q correct to 4 significant figures.

The slant height $l$ of a cone is given by the Pythagorean theorem, where $r$ is the radius and $h_{cone}$ is the height of the cone.

$$l = \sqrt{r^2 + h_{cone}^2}$$

Given $r = 2.1$ m and $h_{cone} = 5$ m.

$$l = \sqrt{(2.1)^2 + (5)^2}$$ $$l = \sqrt{4.41 + 25}$$ $$l = \sqrt{29.41}$$

$$l \approx 5.4231 \text{ m}$$ $$l \approx \boxed{\text{5.42 m}}$$

The total surface area (TSA) of the solid Q is the sum of the surface area of the hemisphere, the lateral surface area of the cylinder, and the lateral surface area of the cone.

$$TSA = (\text{Area of hemisphere}) + (\text{Lateral area of cylinder}) + (\text{Lateral area of cone})$$ $$TSA = 2\pi r^2 + 2\pi r h_c + \pi r l$$

We can factor out $\pi r$:

$$TSA = \pi r (2r + 2h_c + l)$$

Given $r = 2.1$ m, $h_c = 14$ m, $\pi = \frac{22}{7}$, and $l \approx 5.4231$ m (using the unrounded value for accuracy).

$$TSA = \frac{22}{7} \times 2.1 \times (2 \times 2.1 + 2 \times 14 + 5.4231)$$ $$TSA = 6.6 \times (4.2 + 28 + 5.4231)$$ $$TSA = 6.6 \times (37.6231)$$

$$TSA \approx 248.31246 \text{ m}^2$$ $$TSA \approx \boxed{\text{248.3 m}^2}$$

The total volume (V) of the solid Q is the sum of the volume of the hemisphere, the volume of the cylinder, and the volume of the cone.

$$V = (\text{Volume of hemisphere}) + (\text{Volume of cylinder}) + (\text{Volume of cone})$$ $$V = \frac{2}{3}\pi r^3 + \pi r^2 h_c + \frac{1}{3}\pi r^2 h_{cone}$$

We can factor out $\pi r^2$:

$$V = \pi r^2 \left(\frac{2}{3}r + h_c + \frac{1}{3}h_{cone}\right)$$

Given $r = 2.1$ m, $h_c = 14$ m, $h_{cone} = 5$ m, and $\pi = \frac{22}{7}$.

$$V = \frac{22}{7} \times (2.1)^2 \times \left(\frac{2}{3} \times 2.1 + 14 + \frac{1}{3} \times 5\right)$$ $$V = \frac{22}{7} \times 4.41 \times \left(1.4 + 14 + \frac{5}{3}\right)$$ $$V = 13.86 \times \left(15.4 + 1.6666...\right)$$ $$V = 13.86 \times \left(17.0666...\right)$$

$$V \approx 236.5824 \text{ m}^3$$ $$V \approx \boxed{\text{237 m}^3}$$