A simple solvable group must be cyclic of prime order (since it must be abelian, and so cannot have proper [normal] subgroups). But a simple solvable group would not contain nontrivial normal subgroups, so the proposition would be true for such a group by vacuity (the hypotheses are never satisfied). Alternatively, if you allow the whole group to be a "nontrivial normal subgroup", your $H$ can only be $G$ itself, which is already abelian, so you can set $A=H=G$; either way, the proposition is true for such a group.
(If $G$ is solvable, then $[G,G]$ is a proper subgroup of $G$; since it is always normal in $G$, if $G$ is also simple, then we must have $[G,G]=\{1\}$, hence $G$ is abelian).
Hint for the question. If $H$ is abelian, you are done. If not, then $[H,H]$ is nontrivial; use the fact that $H^{(n)}\subseteq G^{(n)}$, where $G^{(k)}$ is the $k$th term of the derived series of $G$, to show that $H$ is solvable, and use the fact that $H\triangleleft G$ to show $[H,H]\triangleleft G$. Then replace $H$ with $[H,H]$, lather, rinse, and repeat.
Different hint. Let $1=G_0\triangleleft G_1\triangleleft\cdots\triangleleft G_s = G$, with $G_i/G_{i+1}$ abelian. By Problem 8, you can pick the $G_i$ normal in $G$. Look at $H_i=H\cap G_i$.
Added. Sigh. I didn't notice that Problem 8 assumes $G$ is finite; the result is true, as witnessed by the derived series, but again it's probably not what you want.
Added 2. Okay, this should work; it takes some of the ideas of the hint in Problem 8 of the same page, so it should be "reasonable". Let $i$ be the largest index such that $G_i\cap H$ is a proper subgroup of $H$. Then $H\subseteq G_{i+1}$, hence $G_i\cap H\triangleleft H$. Moreover, $H/(H\cap G_i) \cong (HG_i)/G_i \leq G_{i+1}/G_i$, so $H/(H\cap G_i)$ is abelian. As in problem 8, this means that $x^{-1}y^{-1}xy\in H\cap G_i$ for all $x,y\in H$. Show that this is also true for all $G$-conjugates of $H\cap G_i$, hence their intersection, which is normal in $G$, contains all $x^{-1}y^{-1}xy$ with $x,y\in H$. If this is trivial, then $H$ is abelian and we are done. If it is not trivial, then this intersection is nontrivial, and normal in $G$. Replace $H$ with this intersection, and note that the largest index $j$ such that the intersection with $G_j$ is proper is striclty smaller than $i$; so you can set up a descent. Lather, rinse, and repeat.
$\langle 1 \rangle$ is already a generator for $\mathbb{Z}$, because a generator also "generates" inverse elements.
This is due to the definition of a generator.
From Wikipedia:
In abstract algebra, a generating set of a group is a subset such that every element of the group can be expressed as the combination (under the group operation) of finitely many elements of the subset and their inverses.
So $-3\in \langle 1 \rangle$, because we have $-3= (-1)+(-1)+(-1)$.
The same goes for $\langle -1\rangle$.
Also note that technically $\langle 1, -1\rangle$ is also a valid generator of the group, but it happens that there is also a "smaller" generator.
Best Answer
As you can see the author is now discussing the subgroups of $G=Z_n$, not $\Bbb{Z}/n\Bbb{Z}$, that's why different notation is used.
In the book, the author denotes $Z_n=\{0,1,\dots,n-1\}$ and $\Bbb{Z}/n\Bbb{Z}=\{\bar{0},\bar{1},\dots,\overline{n-1}\}$