(a) $\dfrac{1}{9}$
Total outcomes = 36. Favourable outcomes (difference = 3): (1,4),(4,1),(2,5),(5,2),(3,6),(6,3) → 4 pairs. P = $\dfrac{4}{36} = \dfrac{1}{9}$.
List pairs where |die1 − die2| = 3: (1,4),(4,1),(2,5),(5,2),(3,6),(6,3) — exactly 4 pairs out of 36. Examiners expect you to identify all favourable outcomes correctly; missing any pair is a common error.