This is really a much longer question than can be easily answered in this format (you should post a paid question without the 20min time limit) - mainly because you've asked four instead of one. Hence why no-one else has answered it I guess :-) But here's a run through of the strategy.
First, you need to convert each of the values you're given in a)-d) into a test statistic. You do this by subtracting it from the mean, and dividing by the standard error of the mean. The standard error of the mean is the standard deviation divided by the square root of your sample size:
SEM = sd/sqrt(n) = 225/sqrt(100) = 22.5
So your test statistics are found by t = (x - 1050)/22.5.
Then you'll compare this to the t-distribution with df = n-1 = 99, to get the associated probability. You have to use the t-distribution here, NOT the normal (z) distribution because you don't know the population standard deviation - only the sample standard deviation. So for example a):
a) t = (1030 - 1050)/22.5 = -0.888888889
P(t<-0.888.., df = 99) = 0.188108765 (use a t-table or an online calculator, or the command =Norm.dist(-0.8888,99,1) in Excel).
For b) follow the same logic through, you should get ~0.134.
For c) find the probabilities for both values, and remember that P(a<t<b) = P(t<b) - P(t<a) since this is "everything to the left of b, that isn't to the left of a", i.e. between the two values. You should get around 0.259.
For d) remember you're looking for >1075, so do the same but find 1 - P(t<...), i.e. everything that's NOT to the left of your t-stat. This should give you ~0.134.
Hope this is helpful!
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