For (1): As Dylan Moreland points out in a comment, the problems lies in (b). It's not true that a rational smaller than an element of your $S$ is also in $S$.
For (2) and (3), I have the same comment in both: A common and understandable confusion when being first exposed to Dedekind cuts is that you have to completely forget that real numbers exist. It doesn't really make sense to say that a cut is on an irrational number, since there are no such thing as rational numbers when you're defining Dedekind cuts. (Of course, most of the things you're saying in this respect are "morally correct" in the sense that they end up being true once you've defined real numbers, but in the long run, this is circular.)
Similarly, for your $2^{1/3}$ Dedekind cut, it may be helpful while you're going through this to never ever use notations like $2^{1/3}$ or $\sqrt[3]{2}$, except possibly as shorthand for the Dedekind cut $S$ you describe (and even then, only once you've defined multiplication of cuts and checked that $S^3=2$), since in the rational world, this number does not exist. So no, you can't just say it's that set, because you don't know that set is a Dedekind cut. You need to prove that it has no maximal element, and if you think about it, checking that there is no rational number whose cube equals 2 is a little shy of that stronger conclusion.
I assume that $A$ is supposed to be a Dedekind cut. Start by making a sketch:
+++++++++++++++++++++++++++++++)----------------------------------
A
The line as a whole represents $\Bbb Q$, and everything to the left of the parenthesis is in the cut $A$. Now look at the definition of $-A$:
$$-A=\{-(q+b):q\in\mathbb{Q}^+, b\in\mathbb{Q}\setminus A\}$$
It depends on two subsets of $\Bbb Q$, $\Bbb Q^+$, and $\Bbb Q\setminus A$, so we should figure out where those are in the picture. $\Bbb Q\setminus A$ is easy:
+++++++++++++++++++++++++++++++)(---------------------------------
A Q\A
it’s everything to the right of $A$. And we know just what $\Bbb Q^+$ is: it’s the set of positive rationals. What happens when you add a positive rational $q$ to every member of $\Bbb Q\setminus A$? You shift the set $\Bbb Q\setminus A$ to the right by $q$ units:
-------------------------------)(+++++++++++++++++++++++++++++++++
A Q\A
ooooooooooooooooooooooooooooooo)-----------(++++++++++++++++++++++
A q + Q\A
That last picture shows $\{q+b:b\in\Bbb Q\setminus A\}$ for a particular $q\in\Bbb Q^+$. Now what happens when you look at the negatives of these rational numbers, $\{-(q+b):b\in\Bbb Q\setminus A\}$? You simply flip the line $180$° around $0$ to get a picture more or less like this:
++++++++++++++++++++++)-----------(ooooooooooooooooooooooooooooooo
The plus signs mark the set $-(q+\Bbb Q\setminus A)$, and the o’s mark the set $\{-a:a\in A\}$. The gap in the middle has length $q$. $\{-a:a\in A\}$ is always in the same place, but the location of $-(q+\Bbb Q\setminus A)$ depends on $q$: when $q$ is large, it’s far to the left of $\{-a:a\in A\}$, and when $q$ is small, it’s very close to $\{-a:a\in A\}$.
The set $-A$ that you’re to prove is a Dedekind cut is just the union of all these sets $-(q+\Bbb Q\setminus A)$ as $q$ ranges over the positive rationals, so it’s the union of all possible sets like those marked with plus signs in the pictures below:
+++++++++++++++++)----------------(ooooooooooooooooooooooooooooooo
++++++++++++++++++++++)-----------(ooooooooooooooooooooooooooooooo
+++++++++++++++++++++++++)--------(ooooooooooooooooooooooooooooooo
++++++++++++++++++++++++++++++)---(ooooooooooooooooooooooooooooooo
What do you think that union will look like in a sketch of this kind? Won’t it look something like this?
+++++++++++++++++++++++++++++++++)(ooooooooooooooooooooooooooooooo
That looks an awful lot like a Dedekind cut. Now you just have to prove it. For instance, you have to show that there is some rational that is not in the set. From the picture it appears that any $-a$ with $a\in A$ should work, so you should try to prove that this is the case. (Remember, the $q$’s that are being added are all strictly positive.)
You need to show that if $p$ is rational and $p<r\in-A$, then $p\in -A$. The picture certainly makes that look plausible, and it’s not hard to show. If $r\in-A$, then $r=-(q+a)$ for some $q\in\Bbb Q^+$ and $a\in A$. Can you find another positive rational $q'$ such that $p=-(q'+a)$? The number $r-p$ may be useful.
Best Answer
Let’s take a closer look at the conditions defining a Dedekind cut. A subset $D$ of $\Bbb Q$ is a Dedekind cut if:
Think of the rationals in the usual pictorial fashion, laid out as a line extending infinitely far in both directions, negative rationals to the left and positive rationals to the right. Condition (1) says that if $D$ contains some rational $r$, it contains every rational to the left of $r$ as well. One set that satisfies this condition is $\Bbb Q$ itself. Another is $(\leftarrow,0)$, the set of negative rationals, containing every rational strictly to the left of $0$. Yet another is $(\leftarrow,1]$, the set of rationals at or to the left of $1$. On the other hand, the set $(\leftarrow,0)\cup(2,4)$ does not satisfy condition (1): it contains $3$, but it doesn’t contain $1$ even though $1<3$. It’s not an initial segment of $\Bbb Q$.
Condition (2) is the simplest: it just says that $D$ cannot extend infinitely far to the right. There must be some rational number $q$ such that every member of $D$ is at or to the left of $q$. Almost all of the sets that satisfy condition (1) satisfy condition (2) as well; the only exception is $\Bbb Q$ itself, and condition (2) is designed specifically to rule out $\Bbb Q$.
Condition (3) is perhaps the hardest to get a grip on, but what it says is actually quite simple: it says that $D$ must not have a largest element. No matter what $r$ you choose in $D$, $D$ contains some bigger number $s$. This rules out sets like $(\leftarrow,1]=\{q\in\Bbb Q:q\le 1\}$, that have a maximum element.
A Dedekind cut, therefore, is a subset of the rational numbers that looks more or less like $(\leftarrow,2)$, say: it’s an initial segment of the rational number line, it’s not the whole line, and it has no largest element. In fact, every $r\in\Bbb Q$ defines a Dedekind cut in just this way, namely, the cut $$(\leftarrow,r)=\{q\in\Bbb Q:q<r\}$$ consisting of every rational to the left of (smaller than) $r$.
However, these aren’t the only Dedekind cuts. For example, let $$D=\{q\in\Bbb Q:q<0\text{ or }q^2<2\}\;.$$ It’s not hard to check that this $D$ satisfies conditions (1)-(3); if you’ve not already seen this, you probably will soon. But $D$ is not $(\leftarrow,r)$ for any rational number $r$, because if it were, we could show that $r^2=2$, when in fact we know that $\sqrt2$ is irrational. It’s these ‘extra’ Dedekind cuts, the ones not of the form $(\leftarrow,r)$ for any rational number $r$, that make Dedekind cuts useful and interesting: they ‘fill in’ the gaps in $\Bbb Q$ corresponding to the irrational numbers and allow us to construct the real numbers rigorously starting with just the rationals. In the end the cuts $(\leftarrow,r)$ for $r\in\Bbb Q$ are going to correspond to the rationals, and the ‘extra’ Dedekind cuts are going to correspond to the irrationals. But in order to make that work, we have to be able to define the various arithmetic operations on these Dedekind cuts in such a way that they behave the way they should. Your exercise here is part of showing how to define multiplication of Dedekind cuts.
I’ll say nothing about the exercise itself, as breeden has already covered that in some detail.
Added: In view of the comments, I think that I should emphasize that in what I’ve written above, $(\leftarrow,q)$ and so forth are to be understood as subsets of $\Bbb Q$. That is, I’m writing $(\leftarrow,q)$ as an abbreviation for $\{r\in\Bbb Q:r<q\}$, not for $\{r\in\Bbb R:r<q\}$.
As another example to show how a set can be bounded above (as is required by (2)) and still have no largest element, let $D$ be the set of negative rationals. Every member of $D$ is less than $0$ (and hence less than every positive rational as well), but $D$ has no largest element. If $m$ and $n$ are positive integers, so that $-\frac{m}n\in D$, then $-\frac{m}n<-\frac{m}{2n}<0$, so $-\frac{m}{2n}$ is a member of $D$ that is bigger than $-\frac{m}n$. This shows that $D$ has no largest element: give me any element of $D$, and I’ve just shown you how to find a bigger one.