This question tests understanding of probability when a die is thrown twice. The hint clarifies that throwing a die twice is equivalent to throwing two dice simultaneously. The total number of possible outcomes is $6 \times 6$ = 36. For part (i), we count outcomes where 5 doesn't appear in either throw. For part (ii), we can either directly count outcomes with at least one 5, or use the complementary probability approach: P(at least once) = 1 - P(never).