1. **Problem statement:** We have triangle ABC with sides |AB| = 5, |AC| = 3, and angles |\angle ABC| = W, |\angle ACB| = 2W, where $0^\circ < W < 45^\circ$. We want to find $\cos W$ in the form $\frac{p}{q}$ with $p,q \in \mathbb{N}$. We will use the result from part (b)(i) and the Sine Rule. 2. **Recall the Sine Rule:** For any triangle, $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$ where $a,b,c$ are sides opposite angles $A,B,C$ respectively. 3. **Label sides and angles:** - Side opposite $W$ (angle at B) is $AC = 3$. - Side opposite $2W$ (angle at C) is $AB = 5$. - Side opposite $\angle A$ is $BC$ (unknown). 4. **Apply the Sine Rule to sides AB and AC:** $$\frac{5}{\sin 2W} = \frac{3}{\sin W}$$ 5. **Rearrange to express $\sin 2W$ in terms of $\sin W$:** $$5 \sin W = 3 \sin 2W$$ 6. **Use the double-angle identity for sine:** $$\sin 2W = 2 \sin W \cos W$$ 7. **Substitute into the equation:** $$5 \sin W = 3 \times 2 \sin W \cos W = 6 \sin W \cos W$$ 8. **Divide both sides by $\sin W$ (nonzero since $0^\circ < W < 45^\circ$):** $$\cancel{\sin W} \times 5 = 6 \cancel{\sin W} \cos W \implies 5 = 6 \cos W$$ 9. **Solve for $\cos W$:** $$\cos W = \frac{5}{6}$$ 10. **Conclusion:** The value of $\cos W$ in the form $\frac{p}{q}$ with $p,q \in \mathbb{N}$ is $$\boxed{\frac{5}{6}}$$ This uses the Sine Rule and the double-angle identity, along with the given side lengths and angles.