Answer:
a) Expected score on the exam is 12.8.
b) Variance 10.24, Standard deviation 3.2
Step-by-step explanation:
For each question, there are only two possible outcomes. Either you guesses the answer correctly, or you does not. The probability of guessing the answer of a question correctly is independent of other questions. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
The expected value of the binomial distribution is:

The variance of the binomial distribution is:

The standard deviation of the binomial distribution is:

64 questions.
So 
5 possible answers, one correctly, chosen at random:
So 
(a) What is your expected score on the exam?

(b) Compute the variance and standard deviation of x. Variance =Standard deviation

Variance 10.24

Standard deviation 3.2