Let the probability that the score is greater than 2 be denoted by \(p\). The score is greater than 2 if the difference between the numbers on the two spinners is 3 or 4. The possible pairs are (1,4), (1,5), (2,5), (4,1), (5,1), (5,2). There are 6 such pairs out of a total of 25 possible outcomes (since each spinner has 5 sides), so \(p = \frac{6}{25}\).

We need to find the probability that the score is greater than 2 on at least 3 occasions out of 9 spins. This is a binomial probability problem where \(X \sim \text{Binomial}(9, \frac{6}{25})\).

The probability that the score is greater than 2 on at least 3 occasions is:

\(P(X \geq 3) = 1 - P(X = 0, 1, 2)\)

Calculate \(P(X = 0, 1, 2)\) using the binomial probability formula:

\(P(X = 0) = \binom{9}{0} \left(\frac{6}{25}\right)^0 \left(\frac{19}{25}\right)^9\)

\(P(X = 1) = \binom{9}{1} \left(\frac{6}{25}\right)^1 \left(\frac{19}{25}\right)^8\)

\(P(X = 2) = \binom{9}{2} \left(\frac{6}{25}\right)^2 \left(\frac{19}{25}\right)^7\)

Substitute and calculate:

\(P(X \geq 3) = 1 - (0.08459 + 0.2404 + 0.3037)\)

\(P(X \geq 3) = 1 - 0.62869 = 0.37131\)

Thus, the probability that the score is greater than 2 on at least 3 occasions is approximately 0.371.