Let's solve the problem based on the normal distribution of scores:

1. For statement (a): The mean score is 52 and the standard deviation is 13. According to the empirical rule (68-95-99.7 rule) for normal distributions:

   • About 68% of the data lies within one standard deviation of the mean. Thus, it's between $\mu - \sigma$ and $\mu + \sigma$.
   • Calculation: $\mu - \sigma = 52 - 13 = 39$ and $\mu + \sigma = 52 + 13 = 65$.
   • Therefore, approximately 68% of the students scored between 39 and 65.

2. For statement (b):

   • To find the percentage of students who scored between 13 and 91, we calculate the z-scores for these values and then refer to the standard normal distribution table.
   • Z-score for 13: $z = \frac{13 - 52}{13} = -3$
   • Z-score for 91: $z = \frac{91 - 52}{13} = 3$
   • According to the standard normal distribution, about 99.7% of the data lies within three standard deviations of the mean. Thus, approximately 99.7% of the students scored between 13 and 91.

Would you like to know more details about any of these calculations, or do you have any other questions? Here are some related questions you might find helpful:

1. What would be the scores for 95% of the students?
2. How would the results change if the standard deviation was larger or smaller?
3. What is the probability of a student scoring above 65?
4. How can these concepts be applied to other distributions?
5. What are the implications of a higher mean score on the student distribution?

Tip: When working with normal distributions, always consider the empirical rule for quick estimations of data spread and probabilities.