The diagram below shows three points A, B and C on level ground. B is on a bearing of N 30°E from point A and ABC = 45°.

Here's the solution to the problem:

a) Step 1: Interpret the bearing N 30°E. This means 30° measured clockwise from the North direction. Step 2: Express this as a three-figure bearing. $\boxed{030^\circ}$

b) (i) Find the bearing of A from B. Step 1: The bearing of B from A is $030^\circ$. Step 2: To find the bearing of A from B (back bearing), add $180^\circ$ to the original bearing since $030^\circ < 180^\circ$. $030^\circ + 180^\circ = 210^\circ$ $\boxed{210^\circ}$

(ii) Find the bearing of B from C. Step 1: The bearing of B from A is $030^\circ$. The bearing of A from B is $210^\circ$. This means that at point B, the line BA makes an angle of $210^\circ$ clockwise from the North line at B. Step 2: We are given $\angle ABC = 45^\circ$. This angle is measured from BA to BC. Since the angle is positive, we assume BC is clockwise from BA. Step 3: The bearing of C from B is the bearing of BA plus $\angle ABC$. $210^\circ + 45^\circ = 255^\circ$ Step 4: To find the bearing of B from C (back bearing), subtract $180^\circ$ from the bearing of C from B since $255^\circ > 180^\circ$. $255^\circ - 180^\circ = 075^\circ$ $\boxed{075^\circ}$

c) Two similar cuboids have corresponding sides in the ratio 4:6.

(i) Express the ratio in its simplest form. Step 1: Divide both parts of the ratio by their greatest common divisor, which is 2. $4:6 = \frac{4}{2}:\frac{6}{2} = 2:3$ $\boxed{2:3}$

(ii) Find the ratio of the volumes of the cuboids. Step 1: If the ratio of corresponding sides of similar figures is $a:b$, then the ratio of their volumes is $a^3:b^3$. Step 2: Using the simplified side ratio $2:3$. $\text{Ratio of volumes} = 2^3:3^3 = 8:27$ $\boxed{8:27}$

(iii) If the volume of the larger cuboid is 81 cm³, find the volume of the smaller cuboid. Step 1: Let $V_S$ be the volume of the smaller cuboid and $V_L$ be the volume of the larger cuboid. From (c)(ii), $\frac{V_S}{V_L} = \frac{8}{27}$. Step 2: Substitute the given volume of the larger cuboid, $V_L = 81 \text{ cm}^3$. $\frac{V_S}{81 \text{ cm}^3} = \frac{8}{27}$ Step 3: Solve for $V_S$. $V_S = \frac{8}{27} \times 81 \text{ cm}^3 = 8 \times 3 \text{ cm}^3 = 24 \text{ cm}^3$ $\boxed{24 \text{ cm}^3}$

(iv) Find the mass of the larger cuboid if its density is 0.5 g/cm³. Step 1: Recall the formula for density: $\text{Density} = \frac{\text{Mass}}{\text{Volume}}$. Therefore, $\text{Mass} = \text{Density} \times \text{Volume}$. Step 2: Use the given density ($0.5 \text{ g/cm}^3$) and the volume of the larger cuboid ($81 \text{ cm}^3$). $\text{Mass} = 0.5 \text{ g/cm}^3 \times 81 \text{ cm}^3 = 40.5 \text{ g}$ $\boxed{40.5 \text{ g}}$