Total number of outcomes = 36

(i) cases where 5 comes up on at least one time are (1,5),(2,5),(3,5),(4,5),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6) and (6,5).

the number of such cases = 11.

The number of cases where up either time = 36 - 11 = 25.

Therefore, P(5 will not come up either time) = \(\frac{number\, of\,favorable\,outcomes}{number\,of \,all\,possible\,outcomes}\) = \(\frac{25}{26}\)

Thus, the probability that 5 will not come up either time is \(\frac{25}{26}\).

(ii) cases where 5 comes up on exactly one time are (1,5),(2,5),(3,5),(4,5),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6) and (6,5).

the number of such cases = 10.

Therefore, P(5 will come up exactly one time) = \(\frac{number\, of\,favorable\,outcomes}{number\,of \,all\,possible\,outcomes}\) = \(\frac{10}{36}\) = \(\frac{5}{18}\)

Thus, the probability that 5 will come up exactly one time is \(\frac{5}{18}\)

(iii) cases, where 5 comes up on exactly two times, is (5,5).

the number of such cases = 1.

Therefore, P(5 will come up both the time) = \(\frac{number\, of\,favorable\,outcomes}{number\,of \,all\,possible\,outcomes}\) = \(\frac{1}{36}\)

Thus, the probability that 5 will come up both the time is \(\frac{1}{36}\).