Apopulation, with an unknown distribution, has a mean of 80 and a standard deviation of 7. for a sample of 49, the probability that the sample mean will be larger than 82 is

Since the distribution is unknown, we have to think of the CLT (Central Limit Theorem):
n = 49

μ(x) = μ →→ μ(x) = μ = 80

σ(x) = σ/(√v) →→ 7/√49 = 7/7 = 1

Z(x) = (X-μ)/[σ(x)]

Z(x) = (82-80) /1
Z(x) = - 2
For Z= - 2, the area (probability) = -0.0228 (from the left)
and due to symmetry this area  is equal in absolute value to the sample larger than 82 (to the right), hence the P(X>82) = 0.228

Ican't do any of the other stuff, but i can get you started. the answer, i believe, is $2.50.