= Solution
A <finitely presented group> is one isomorphic to $F(X)/\langle\!\langle R\rangle\!\rangle$, where $X$ and $R$ are finite, $F(X)$ is the <free group> on $X$, and the double brackets mean <normal closure>. Equivalently, there is a surjection from a free group of finite rank onto $G$ whose kernel is normally generated by finitely many words. The universal property of the <free group> extends choices of generator images uniquely, and quotienting imposes the chosen relations.
For the <semidirect product>, choose <finite group presentations>
$$
N=\langle a_1,\ldots,a_r\mid R\rangle,\qquad H=\langle b_1,\ldots,b_s\mid S\rangle.
$$
For each $i,j$, choose an $a$-word $w_{ji}$ representing $\phi(b_j)(a_i)$. We claim that a <finite presentation of a semidirect product> is
$$
\boxed{N\rtimes_\phi H\cong\langle a_1,\ldots,a_r,b_1,\ldots,b_s\mid R,S,\ b_ja_ib_j^{-1}=w_{ji}\ (1\le i\le r,1\le j\le s)\rangle.}
$$
All relations hold in the desired <semidirect product>, so the presented group maps onto it. We must also show there are no extra elements in the kernel. For inverse conjugations, choose an $a$-word $v_{ji}$ representing $\phi(b_j)^{-1}(a_i)$. Conjugating this word by $b_j$ using the displayed relations gives an $a$-word representing $\phi(b_j)\phi(b_j)^{-1}(a_i)=a_i$, so $R$ proves $b_j^{-1}a_ib_j=v_{ji}$. Thus we can move both positive and negative $b$-letters to the right, rewriting every word as $u(a)v(b)$. If its image in $N\rtimes H$ is trivial, projection to $H$ makes $v$ trivial using $S$; its remaining image in $N$ then makes $u$ trivial using $R$. The kernel is trivial, proving the claim and finite presentability.
For a <poly-infinite cyclic group>, write its subnormal series as $G=G_0\triangleright G_1\triangleright\cdots\triangleright G_n=1$ with infinite cyclic factors. Given a subgroup $K\le G$, intersect the series with $K$. The homomorphism $K\cap G_i\to G_i/G_{i+1}$ has kernel $K\cap G_{i+1}$, so
$$
(K\cap G_i)/(K\cap G_{i+1})\hookrightarrow G_i/G_{i+1}\cong\mathbb Z.
$$
Each factor is trivial or infinite cyclic; deleting repetitions gives a poly-infinite cyclic series for $K$. For an extension $1\to N\to E\to Q\to1$ with $N,Q$ <poly-infinite cyclic groups>, lift the series of $Q$ to its full preimages in $E$ and append the series of $N$. The lifted quotient factors are those of $Q$, and the lower factors are those of $N$. Normality at every consecutive step is preserved. This proves closure under both subgroups and extensions.
Finite presentability follows by induction on the length of a poly-infinite cyclic series. The trivial group is finitely presented. For a nontrivial group, its first proper term $G_1$ has a shorter series and $G/G_1\cong\mathbb Z$. Lift $1\in\mathbb Z$ to an element $t\in G$. Its powers give an injective section of the quotient map, since $t^k$ maps to $k$. Hence $G\cong G_1\rtimes\langle t\rangle$. By induction $G_1$ is finitely presented, and $\langle t\rangle\cong\mathbb Z$ has a finite presentation. The <finite presentation of a semidirect product> proved above completes the induction. Thus \b[every poly-infinite cyclic group is finitely presented].
For the final example, take the <dyadic affine Baumslag-Solitar group>
$$
G=\langle a,t\mid tat^{-1}=a^2\rangle\cong\mathbb Z[1/2]\rtimes\mathbb Z,
$$
where the generator of $\mathbb Z$ acts on the additive group $\mathbb Z[1/2]$ by multiplication by two. Here is a verification of this identification, rather than an assumption about the presentation. Put $a_i=t^{-i}at^i$ for $i\ge0$. The relation gives $a_{i+1}^2=a_i$, so
$$
\langle a_0\rangle\le\langle a_1\rangle\le\cdots.
$$
Their union $A$ is abelian and is normal: conjugation by $t^{-1}$ shifts $a_i$ to $a_{i+1}$, while conjugation by $t$ shifts $a_i$ to $a_{i-1}$ for $i\ge1$ and sends $a_0$ to $a_0^2$. Every word can consequently be written $a_i^m t^k$ for some $i\ge0$ and $m,k\in\mathbb Z$.
Let $a$ act on the rational line by $x\mapsto x+1$ and $t$ by $x\mapsto2x$. These affine transformations satisfy the relation, and $a_i^m t^k$ acts by $x\mapsto2^kx+m/2^i$. It is the identity only when $k=m=0$. Thus the action is faithful and identifies $A$ with all dyadic translations. We have the required series
$$
\boxed{G\triangleright\mathbb Z[1/2]\triangleright1,\qquad G/\mathbb Z[1/2]\cong\mathbb Z.}
$$
Both factors are <torsion-free abelian groups>, and $G$ is visibly finitely presented. But $\mathbb Z[1/2]$ is not finitely generated, since finite sets have bounded powers of two in their denominators. Every <poly-infinite cyclic group> has the <maximal condition on subgroups>: its factors $\mathbb Z$ have that condition, and Question 1 proves closure under extensions. Therefore $G$ cannot be poly-infinite cyclic. This gives \b[a finitely presented counterexample to replacing infinite cyclic factors by arbitrary torsion-free abelian factors].
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