It depends very much on the enumeration $(r_n)$.
If $(r_n)$ is an enumeration obtained by choosing $r_{n+1}$ in the interval $(r_n+1/(2n),r_n+1/n)$ if $r_n\lt1-1/(2n)$, and in the interval $(0,1/n)$ otherwise, then the answer is yes.
If $(r_{6^n})$ enumerates the rational numbers in $[0,\frac12]$ and $(r_n)$, $n$ not a power of $6$, enumerates the rational numbers in $(\frac12,1]$, then $[0,\frac12]$ is not entirely covered hence the answer is no.
The authors write $N\approx c|z|!$ Thus there won't be one $N$ that works for all $z.$
Given $z,$ choose $N$ such that $|z|/t\le N\le |z|/t+1.$ Then
$$|F_2(z)| \le \prod_{n=N+1}^{\infty}(1+e^{-2\pi nt}e^{2\pi |z|})$$ $$ \le \exp \left ( \sum_{N+1}^{\infty}e^{-2\pi nt}e^{2\pi |z|}\right )$$ $$ = \exp \left( e^{-2\pi (N+1)t}e^{2\pi |z|}/(1-e^{-2\pi t})\right)\le \exp (1/(1-e^{-2\pi t})).$$
Thus we have a bound on $|F_2(z)|$ independent of $z.$
Added later For the other estimate, let's start with $|F_1(z)|\le 2^Ne^{2\pi N|z|}.$ Now
$$2^Ne^{2\pi N|z|} < e^Ne^{2\pi N|z|} \le e^{|z|+1}e^{2\pi (|z|+1)|z|},$$
where we have used $N\le |z|+1.$ The expression on the right has the form $e^{a+b|z|+c|z|^2},$ where $a,b,c$ are positive constants. This is not bounded above by $e^{c'|z|^2}$ for any $c'>0.$ The problem is with $|z|$ small, not $|z|$ large (the inequality falls apart when $|z|=0$). So they made a mistake in the hint there, but it's no problem. Remember, the ultimate aim is to show $|F_1(z)|\le Ae^{\alpha |z|^2}$ for some positive $A,\alpha.$ Writing $A=e^\beta,$ we want
$$e^{a+b|z|+c|z|^2}\le e^{\beta +\alpha |z|^2}$$
for some choice of $\alpha,\beta >0.$ That's easy: Let $\alpha = b+c,\beta =a+b+c.$ (Think about $|z|\le 1, |z|\ge 1$ separately.)
Best Answer
Fix some irrational $\alpha$ and any enumeration $\{q_n:n\in\Bbb Z^+\}$ of the rationals. We’ll build a new enumeration $\{p_n:n\in\Bbb Z^+\}$ of $\Bbb Q$ in such a way that for each $n\in\Bbb Z^+$, $\alpha\notin(p_n-\frac1n,p_n+\frac1n)$.
Let $n_1=\min\{n\in\Bbb Z^+:|q_n-\alpha|\ge 1\}$, and let $p_1=q_{n_1}$; clearly $\alpha\notin(p_1-1,p_1+1)$. Let $Z_1=\Bbb Z^+\setminus\{n_1\}$, the set of indices of rationals not yet re-enumerated.
Now suppose that we’ve already defined $p_1,\dots,p_m$ and $Z_m$. Let $$n_{m+1}=\min\left\{n\in Z_m:|q_n-\alpha|\ge\frac1{m+1}\right\}\;,$$ and set $p_{m+1}=q_{n_{m+1}}$ and $Z_{m+1}=Z_m\setminus\{n_{m+1}\}$. Clearly $p_{m+1}$ is distinct from $p_1,\dots,p_m$ and $$\alpha\notin\left(p_n-\frac1n,\,p_n+\frac1n\right)\;.$$
All that’s left is to prove that every rational is eventually enumerated as $p_n$ for some $n\in\Bbb Z^+$. That follows from the fact that at each stage we took the first available rational in the original enumeration; I’ll leave it to you to fill in the details, unless you get stuck and ask for help.