Okello has an ice cream bucket in the shape of a frustum... How much money did Okello earn?

The bucket is a frustum with top diameter 48 cm (radius $R = 24$ cm), bottom diameter 14 cm (radius $r = 7$ cm), and height $H = 20$ cm.
The volume of a frustum is given by $V = \frac{1}{3}\pi H (R^2 + Rr + r^2)$.

$$V_{bucket} = \frac{1}{3}\pi (20 \text{ cm}) ((24 \text{ cm})^2 + (24 \text{ cm})(7 \text{ cm}) + (7 \text{ cm})^2)$$ $$V_{bucket} = \frac{20\pi}{3} (576 + 168 + 49) \text{ cm}^3$$ $$V_{bucket} = \frac{20\pi}{3} (793) \text{ cm}^3 = \frac{15860\pi}{3} \text{ cm}^3$$

Each tin is a cylinder with diameter 7 cm (radius $r_{tin} = 3.5$ cm) and height $h_{tin} = 5$ cm.
The volume of a cylinder is given by $V = \pi r^2 h$.

$$V_{tin} = \pi (3.5 \text{ cm})^2 (5 \text{ cm})$$ $$V_{tin} = \pi (12.25) (5) \text{ cm}^3 = 61.25\pi \text{ cm}^3$$

The number of tins is the total volume of ice cream divided by the volume of one tin. We take the floor since only full tins are sold.

$$N = \frac{V_{bucket}}{V_{tin}} = \frac{\frac{15860\pi}{3} \text{ cm}^3}{61.25\pi \text{ cm}^3}$$ $$N = \frac{15860}{3 \times 61.25} = \frac{15860}{183.75} \approx 86.31$$

So, 86 full tins can be filled.

Each tin is sold for UGX 2,000.

$$\text{Total money} = 86 \times \text{UGX } 2,000 = \text{UGX } 172,000$$

Okello earned $\boxed{\text{UGX 172,000}}$.

b) To determine how close Okello moved towards the tree, we use trigonometry to find his initial and final distances from the tree.

The height of the tree is 25 m, and Okello's height is 1.5 m. The effective height for the angle of elevation is the height of the mango above Okello's eye level.

$$H_{eff} = 25 \text{ m} - 1.5 \text{ m} = 23.5 \text{ m}$$

The initial angle of elevation is $32^\circ$. Let $D_1$ be the initial horizontal distance.

$$\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}$$ $$\tan(32^\circ) = \frac{23.5 \text{ m}}{D_1}$$ $$D_1 = \frac{23.5 \text{ m}}{\tan(32^\circ)} \approx \frac{23.5 \text{ m}}{0.62487} \approx 37.607 \text{ m}$$

The angle of elevation increased by $18^\circ$, so the final angle $\theta_2 = 32^\circ + 18^\circ = 50^\circ$. Let $D_2$ be the final horizontal distance.

$$\tan(50^\circ) = \frac{23.5 \text{ m}}{D_2}$$ $$D_2 = \frac{23.5 \text{ m}}{\tan(50^\circ)} \approx \frac{23.5 \text{ m}}{1.19175} \approx 19.719 \text{ m}$$

The distance Okello moved is the difference between the initial and final distances.

$$\text{Distance moved} = D_1 - D_2 \approx 37.607 \text{ m} - 19.719 \text{ m} \approx 17.888 \text{ m}$$

Okello moved approximately $\boxed{\text{17.89 m}}$ closer to the tree.