(a) $\dfrac{7}{36}$
Total outcomes = 36. Favourable outcomes (sum = 2, 3, or 5): sum 2 → (1,1); sum 3 → (1,2),(2,1); sum 5 → (1,4),(2,3),(3,2),(4,1) → 7 outcomes. P = $\dfrac{7}{36}$.
Count outcomes for each sum separately: sum of 2 gives 1 outcome, sum of 3 gives 2 outcomes, sum of 5 gives 4 outcomes — total 7. Divide by 36 (total outcomes when two dice are rolled). Students often miss that sum = 4 is not asked, so don't include those pairs.