Looks like you got "a" correct but "b" incorrect.

We have to find B) and C) first in order to find A)!!

z³+az+b = 0

z³+0z²+az+b = 0


There are 3 roots to a 3rd degree equation:

Since the coefficient of z³ is 1, and the
coefficient of z² is 0,

the sum of the roots is 0.  <-- answer to C)

Since 1+5i is a root, and since all coefficients are
real, 1-5i is also a root.

Let the 3rd root be c

Sum of roots = (1+5i)+(1-5i)+c = 0
                           2+c = 0
                             c = -2

So the three roots are 1+5i, 1-5i, and -2  <-- answer to B).

We substitute -2 for z in the original:

      z³+az+b = 0
(-2)³+a(-2)+b = 0
      -8-2a+b = 0

Since the constant term b is the product of the roots
with the opposite sign,

Product of roots = -(1+5i)(1-5i)(-2) = b
                       -(1-25i²)(-2) = b
                     -[1-25(-1)](-2) = b
                         -[1+25](-2) = b
                             -26(-2) = b
                                  52 = b

We substitute b = 52 in -8a-2a+b = 0:

      -8-2a+b = 0
     -8-2a+52 = 0
       -2a+44 = 0
          -2a = -44
            a = 22 

We have already answered B) and C)

Edwin