Find a value of the standard normal random variable z, call it z0, such that
a. p(z ≤ z0) = .0401
b. p(- z0 ≤ z ≤ z0) = .95
c. p(- z0 ≤ z ≤ z0) = .90
d. p(- z0 ≤ z ≤ 0) = .2967

the 0 is subzero, don't know why the site couldn't show it.

For the answer to this questions,a. P (z ≤ z0) = 0.0401=P(z ≥ z0) = 1-0.0401 = 0.9599 = P(z ≤ -z0) = 0.9599 
From tables z0 = -1.75 

b. P (-z0 ≤ z ≤ z0) = .95 = P (z ≤ z0)- P (z ≤ -z0) = P (z ≤ z0)- P (z ≥ z0) = 
P (z ≤ z0)-(1- P (z ≤ z0)) 
P (z ≤ z0) = (0.95+1)/2=0.975 
From tables z0 = 1.96 

c. P (-z0 ≤ z ≤ z0) = 0.90 
the procedure is the same that exercise b P (z ≤ z0) = (0.9+1)/2=0.95 
From tables the nearest value is z0 = 1.64 

d. P (-z0 ≤ z ≤ 0) = 0.2967= P (z ≤ 0) - P (z ≤ -z0) = P (z ≤ 0) - P (z ≥ z0) = 
P (z ≤ 0) - (1- P (z ≤ z0)) 
P (z ≤ z0) = 0.2967 + 1 - P (z ≤ 0)= 0.2967 + 1 - 0.5 = 0.7967 
From tables z0 = 0.83 

I hope my answer helped you