Let X be the random variable number of people who can't serve.
Why not just calculate Pr(X = 8) + Pr(X = 9) + Pr(X = 10) + Pr(X = 11) + Pr(X = 12) .......? Less to calculate than how I think you've tried to do it .....
b) Let Y be the random variable number of people who can serve.
Then Y ~ Binomial(n = ?, p = 3/4).
You require the smallest integer value of n such that .
The easiest approach to solving this inequality is to get the technology available to you to do trial-and-error - perhaps setting up a table of probabilities with n.
Using my TI-89 and defining Y1 = binomcdf(x, 0.75, 12, x) and using a table of values for Y1, I easily get that n = 20 ........
Certainly I think it will be a hopeless task trying to solve it without using technology that lets you easily calculate the cumulative probability => you ought to have access to such technology => you should apply that technology to part a ......... !