This is a binomial distribution problem, with the probability of success p = .90

If the number of trials n = 10, the mean number of successes is np = 10 * .90 = 9.

The probability that he makes at least 8 free throws which is P(makes 8 and misses 2) + P(makes 9 and misses 1) + P(makes 10 and misses 0).

This is C(10, 8).9^{8}.1^{2} + C(10, 9).9^{9}.1^{1} + C(10,10).9^{10}.1^{0} where C(10, 8) is the number of combinations of 10 objects taken 8 at a time [sometime written as 10C8], etc. Numerically, this is .1937 + .3874 + .3487 = .9298

One could also get this result by calculating cummulative binomial distribution Binomcdf( trials = 10, p = .9, x value = 7) which is the cummulative probability of 0 up to and including 7 successes out of 10 trials, and subtracting this number from 1.

The second part is a little more abstract. We are redefining success to be the probability of getting at least 8 successes out of 10, and have calculated this to be .9298. In this case, p = .9298, and for n = 10, our expected value is np = 9.298. This, means, out of a several runs of 10 tries, the player will on average get 9.298 successes. The variance is npq, where q is 1-p. So the variance is 10*.9*.1 = .9, and the standard deviation is √variance = √.9 = .9487