(a) The probability of not obtaining a 6 on a single throw is \(\frac{5}{6}\). The probability of not obtaining a 6 for 8 consecutive throws is \(\left( \frac{5}{6} \right)^8\). Therefore, the probability that obtaining a 6 takes more than 8 throws is \(\left( \frac{5}{6} \right)^8 = 0.233\).

(b) For two dice, the probability of obtaining a pair of 6s on a single throw is \(\frac{1}{36}\). The expected number of throws to get a pair of 6s is the reciprocal of this probability, which is \(\frac{1}{\frac{1}{36}} = 36\).

(c) The probability of obtaining a pair of 6s on the 10th throw is \(\left( \frac{35}{36} \right)^9 \times \frac{1}{36}\). The probability of obtaining a pair of 6s on the 11th throw is \(\left( \frac{35}{36} \right)^{10} \times \frac{1}{36}\). Therefore, the probability that obtaining a pair of 6s takes either 10 or 11 throws is \(\left( \frac{35}{36} \right)^9 \times \frac{1}{36} + \left( \frac{35}{36} \right)^{10} \times \frac{1}{36} = 0.0425\).