A polynomial of degree 3 has exactly 3 solutions if we use complex numbers. If the polynomial has real coefficients and z is a complex solution, i.e., [tex]p(z)=0[/tex], the conjugate is also a solution: [tex]p(\overline{z})=0[/tex]
So, in the first case, if we know that 1-i is a solution, the conjugate 1+i must be a solution as well.
So, we know all the roots: -5, 1+i, 1-i.
When you know the solutions [tex]x_1,\ x_2,\ldots,\ x_n[/tex] of a polynomial, you can write the polynomial (up to multiples) as
[tex]p=(x-x_1)(x-x_2)\ldots(x-x_n)[/tex]
So, in your case, we have
[tex]p(x) = (x+5)(x-1-i)(x-1+i) = x^3+3x^2-8x+10[/tex]
You can check that indeed [tex]p(0)=10[/tex], because the constant term is 10, so, we're ok with this
The second exercise is exactly the same: the solutions -6, i, -i identify the polynomial
[tex]p(x)=(x+6)(x-i)(x+i)=x^3+6x^2+x+6[/tex]
In this case, P(-3)=30, so if we want P(-3)=60 we have to multiply everything by 2:
[tex]p(x)=2x^3+12x^2+2x+12[/tex]
