Here's some hints/thoughts.

3.24: This one isn't actually too bad, just gets a little confusing because of the square root and the extra variable k. First note that as long as $x \geq 0$, solving $x + 2\sqrt{x} + k = 0$ is basically solving $u^2 + 2u + k = 0$ where $u = \sqrt{x}$. We make this substitution so that we can use our usual tricks and tools for quadratics from here. Using the quadratic formula we can actually solve to get that $u = -1 \pm \sqrt{1-k}$. Therefore, $\sqrt{x} = -1 \pm \sqrt{1-k}$. This allows us to solve for $x$, just be careful to make sure that we make sure that $x\geq 0$!

For absolute values, you can always break a problem down into cases if needed. Sometimes things get messy that way, but if you're stuck, that can be a good way to start to approach the problem.

6.22: There's a little bit of a trick we can do here, but to start, just try looking at the cases. What happens if $x\geq -2$? What happens if $x < -2$? Note in both cases you should be able to simplify the equation and get rid of all the absolute values. For example, if $x \geq -2$, then we know that $x+2$ is positive so $|x+2| = x+2$.

6.27: You can do a trick here by making the substitution $z = y^2+y$ to simplify things. After this substitution you should get the simpler equation $|z| = 2z+2$. From here we can do cases of $z\geq 0$ and $z < 0$ like in 6.22.

6.30: This is a 3 part question. It should be fairly clear that any value of $x$ is okay to plug in, so the domain is all real numbers. For the range, remember that absolute values are never negative. Therefore, the minimum value of something like $y = |\cdots| - 1$ is often $-1$. Try to see if that makes sense for this situation as well. Finally, to find the zeros, a few different cases could be used. One way is to try working your way "inside" out here. For example,$$||x+2|-3| - 1 = |(x+2) - 3| - 1 = |x-1| - 1 \text{ if } x \geq -2,$$but$$||x+2|-3|-1 = |-(x+2)-3| - 1 = |-x-5| - 1 \text{ if } x < -2.$$

Hope this helps a bit!