The probability of an event is the number of outcomes favorable to the event divided by the total number of possible outcomes. In this case, the event is “the sum is neither 8 nor 9.” There are 36 possible outcomes when a die is thrown twice, and 10 of those outcomes are favorable to the event (namely, (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)). Therefore, the probability of the event is $\frac{10}{36} = \boxed{\frac{5}{18}}$.
Option A is incorrect because it is the probability of the sum being 8. Option B is incorrect because it is the probability of the sum being 9. Option C is incorrect because it is the probability of the sum being 1, 2, 3, 4, 5, or 6. Option D is incorrect because it is the probability of the sum being anything other than 7.