= Solution
The <maximal condition on subgroups>, denoted $\mathrm{max}$, says that every ascending sequence $H_1\le H_2\le\cdots$ of <subgroups> eventually stabilizes. Suppose a subgroup $H$ were not finitely generated. Starting with the trivial subgroup, repeatedly choose an element of $H$ outside the subgroup generated so far. This produces a strictly ascending sequence, contradicting $\mathrm{max}$.
Conversely, suppose every subgroup is a <finitely generated group>. The union $H=\bigcup_iH_i$ of an ascending sequence is a subgroup. A finite generating set of $H$ lies in a single $H_m$, since the sequence is ascending. Thus $H=H_m$, and $H_i=H_m$ for all $i\ge m$. We have proved
$$
\boxed{G\text{ has }\mathrm{max}\quad\Longleftrightarrow\quad\text{every subgroup of }G\text{ is finitely generated}.}
$$
For the extension arguments, let $N\triangleleft G$ and let $\pi:G\to G/N$ be the quotient map. If $N$ and $G/N$ are <finitely generated groups>, choose generators $a_1,\ldots,a_r$ of $N$ and lifts $b_1,\ldots,b_s$ of generators of $G/N$. For every $g\in G$, its image is a word $w(\pi(b_1),\ldots,\pi(b_s))$, so $w(b_1,\ldots,b_s)^{-1}g\in N$. Therefore
$$
G=\langle a_1,\ldots,a_r,b_1,\ldots,b_s\rangle,\qquad d(G)\le d(N)+d(G/N).
$$
Here $d$ denotes the <minimal number of generators>. If instead $N$ and $G/N$ satisfy the <maximal condition on subgroups>, take any $H\le G$. Its intersection $H\cap N$ and its image $\pi(H)$ are finitely generated by the condition in those two groups. The exact sequence $1\to H\cap N\to H\to\pi(H)\to1$ and the preceding argument make $H$ finitely generated. The equivalence already proved now gives $\mathrm{max}$ for $G$. Hence \b[both finite generation and the maximal condition are preserved by <group extensions>].
An infinitely generated group of finite <Prüfer rank> is the additive group $\mathbb Q$. A finite set of rational numbers lies in $D^{-1}\mathbb Z$ for a common denominator $D$, and its generated subgroup is cyclic: after multiplying by $D$, use the <greatest common divisor> of the integer numerators. Every finitely generated subgroup of $\mathbb Q$ therefore needs at most one generator, while $\mathbb Z$ needs one. But no finitely generated subgroup contains $1/(2D)$ once its elements all lie in $D^{-1}\mathbb Z$. Thus
$$
\boxed{\mathbb Q\text{ is not finitely generated and has Prüfer rank }1.}
$$
If $G$ has <Prüfer rank> $r$, every finitely generated subgroup of any subgroup $K\le G$ is also a finitely generated subgroup of $G$, so $\operatorname{rk}_{\mathrm P}(K)\le r$. For a quotient $G/N$, let $A=\langle\bar g_1,\ldots,\bar g_k\rangle$ be finitely generated and choose lifts $g_i\in G$. The subgroup $L=\langle g_1,\ldots,g_k\rangle$ needs at most $r$ generators and maps onto $A$, so $d(A)\le d(L)\le r$. This proves the rank bound for every subgroup and quotient.
Now suppose $N$ and $G/N$ have <Prüfer ranks> $r$ and $s$. For finitely generated $H\le G$ with $H\cap N$ also finitely generated, the extension generating argument gives
$$
d(H)\le d(H\cap N)+d(\pi(H))\le r+s.
$$
The final request needs an argument even when $H\cap N$ is not finitely generated. Write $H=\langle h_1,\ldots,h_k\rangle$. Choose at most $s$ elements $x_1,\ldots,x_s\in H$ whose images generate $\pi(H)$, omitting unused entries if fewer are needed. For each $h_i$, choose a word $w_i$ in these $x_j$ with the same image and set
$$
n_i=w_i(x_1,\ldots,x_s)^{-1}h_i\in H\cap N.
$$
The finitely generated subgroup $K=\langle n_1,\ldots,n_k\rangle$ of $N$ has generators $y_1,\ldots,y_r$, again allowing fewer. Each original $h_i$ belongs to $\langle x_1,\ldots,x_s,y_1,\ldots,y_r\rangle$, and all these new generators belong to $H$. Thus $d(H)\le r+s$ without requiring finite generation of the full intersection. Taking the supremum proves the <Prüfer rank bound for extensions>:
$$
\boxed{\operatorname{rk}_{\mathrm P}(G)\le r+s.}
$$
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