The total probability should sum to 1, so:

f(0)+f(1)+f(2)+f(3)=1

so:

c{(4)+(5)+(8)+(13)}=1

Which you should be able to solve.

mean=sum[ n f(n), n=0,1,2,3]b) Find the mean and variance of X.

variance=sum[ (n-mean)^2 f(n), n=0,1,2,3]

so you are left with the arithmetic.

RonL