For 9.13, you are correct. This should now be fixed.

For the more:

9.21:  If $r > s$ are the roots, we know $r+s = -p$, $rs=1$, and $r-s = 1$. Can you use some of these to solve for $r$ and $s$?

9.23:  The "using Vieta's" part comes from setting up the cubic you need to solve, not solving the cubic (which you need to do by some brute force).

For 9.6 and 9.26, try making the substitution $u = x+y$ and $v = xy$.

For 9.27:  In part (a) you should have $P(z) = 1\cdot z^3 + 0\cdot z^2 + (-3xy)\cdot z + (x^3+y^3)$. Hence in part (b) you want to show $P( -(x+y) ) = 0 $ (that is, substituting $z = -(x+y)$ simplifies to $0$) so $-(x+y)$ is a "zero" of the polynomial $P(z)$.

9.28 is tricky. As a hint, we know $t = -(a+b)(b+c)(a+c)$ (be sure you can explain why). Compare this to the expression $(ab+bc+ac)(a+b+c)$. (Remember we are trying to write everything as expressions we know using Vieta's.)

For 9.29:  It is easy to calculate $x +y = 6$ and $xy = 4$. Use this to your advantage.

Hope these help!