The answer to question 1 is yes.

Pick any field element $x_1$ outside the rationals. Let $x_2$ and $x_3$ be its conjugates under the Galois group. Then the cross ratio $$w=\frac{(x_1-1)(x_2-x_3)}{(x_1-x_2)(1-x_3)}$$ does the trick. (This is the same argument as in this answer.)

Note that these $w$ usually won't be algebraic integers, and they cannot be algebraic integers when the prime $2$ splits in $K$ (since $w$ and $1-w$ cannot both map to $1$ in the 2-element residue class field).

The simplest answer to question 3 I know of is that the set is what Daniel Shanks studied in The Simplest Cubic Fields. (This is a little white lie: Shanks had in fact started out by calling all cyclic cubic fields "simplest", and then proceeded to focus on a certain subset where the conductor is prime.) If $w$ is an algebraic unit in a cubic field satisfying the two norm conditions, then the minimal polynomial of $-w$ is necessarily of the Shanks form $t^3-at^2 -(a+3)t-1$ for some rational integer $a$, and passing to $1/w$ if necessary one can assume $a\ge -1$. So you can enumerate all "green" fields by letting $a$ vary (with only a few repetitions, but that's a deep result) and computing conductors (since $w$ won't necessarily generate the full ring of integers: $a=1259$ being the most spectacular case).

I can't answer question 2 offhand. I rather think "green" fields are rare, but I'm insufficiently familiar with fields that *don't* contain exceptional units to have a good intuition. The possible conductors can of course be enumerated, and a lower bound for $G_N$ follows from the observation that the above family of polynomials yields conductors which must divide $a^2 +3a+9$. It is clear that those $a$ for which this quantity is prime (it will then equal the conductor) contribute a negligible number of "green" fields - there are far too many primes congruent to 1 mod 6. But I don't know how much those $a$ for which the conductor properly divides this quantity will affect the count.

For fun, here's a little table; I've checked this against my archives and it is complete as far as it goes (skipping no "green" fields apart from the gaps indicated by vertical dots). The first four entries cover all cases where multiple $a$ parameters give the same $f$:
$$\begin{array}{r|l|l}
f & a & \text{reduced polynomial} \\ \hline
7 & \{-1,5,12,1259\} & \\
9 & \{0,3,54\} & \\
13 & \{1,66\} & \\
19 & \{2,2389\} & \\
31 & \text{---} & \\
37 & \{4\} & \\
43 & \text{---} & \\
61 & \{39\} & \\
9\cdot 7 & \{6\} & x^3-21x-35 \\
9\cdot 7 & \text{---} & x^3-21x-28 \\
\ldots & \text{---} & \\
79 & \{7\} & \\
\ldots & \text{---} & \\
97 & \{8\} & \\
\ldots & \text{---} & \\
9\cdot 13 & \{9\} & x^3-39x-91 \\
\ldots & \text{---} & \\
139 & \{10\} & \\
\vdots & \vdots & \vdots \\
7\cdot 31 & \{201\} & x^3-x^2-72x+225 \\
7\cdot 31 & \{13\} & x^3-x^2-72x-209 \\
\ldots & \text{---} & \\
241 & \{286\} & \\
13\cdot 19 & \{14\} & x^3-x^2-82x+311 \\
\vdots & \vdots & \vdots \\
373 & \{1598\} & \\
379 & \{911\} & \\
9\cdot 43 & \text{---} & x^3-129x-215 \\
9\cdot 43 & \{18\} & x^3-129x-559
\end{array}$$

Some further remarks:

Cyclic cubic fields of conductor $f>7$ do not contain any units $w$ of positive norm such that $1-w$ is also a unit. (This is due to Nagell, and elementary.)

When $x$ is a totally positive algebraic integer in a totally real number field, $x+1$ can never be an algebraic unit.

When $x$ is a proper power of a nonzero algebraic integer in a totally real number field, $x+1$ can never be an algebraic unit. (Because the norm of $x+1$ can then be expressed as a resultant which also equals the norm of an obvious non-unit in a cyclotomic field.)

The roots of the Shanks polynomials always generate the subgroup of units of positive norm of the ring generated by these roots. (Due to Thomas 1979, based on geometry of numbers results of Berwick 1932, though Ljunggren may have been aware of this fact by 1942.)

Marie-Nicole Gras conjectured in 1976 that whenever a Galois-invariant subgroup of units of positive norm in a cyclic cubic field contains a unit $x$ such that $x+1 \in \mathcal{O}_K^\times$, the same is true already of a generator (as a Galois module) of this subgroup. In other words, the subgroup itself is then already generated by the roots of a Shanks polynomial. This is wide open, but true in a large range.

There are a lot more congruence obstructions beside the obvious one modulo 2. If $x$ is a unit of positive norm in a cyclic cubic field and $x^\sigma$ one of its conjugates, with common minimal polynomial $x^3-tx^2+s-1$, and $U$ the subgroup of units they generate, necessary conditions for $U+1$ intersecting the set of units of negative norm of $K$ nontrivially include:

- $\mathrm{Norm}_{K/\mathbb{Q}}(x+1) \equiv -1 \pmod{2f}$
- (Corollary: $U^{\sigma-1}+1$ meets $\mathcal{O}_K^\times$ only when $f=9$)
- $\mathrm{gcd}(s+1,t+1) = 1$
- $\mathrm{gcd}(s,t) \in \{1,3\}$ unless $f=9$
- $\mathrm{gcd}(s-1,t-1) \in \{1,5\}$
- $\mathrm{gcd}(s-2,t-2) \in \{1,7\}$
- $\mathrm{gcd}(s-3,t-3) \in \{1,3,9\}$

(M.-N. Gras 1973, 1976 with a correction by yours truly to the last item where the $9$ had been missing).

Yet more obstructions arise when $U$ maps to $1$ or onto a too-small multiplicative subgroup in some residue class field. A fun example is $f=211$, where one can work modulo 199.

Selected references:

M.-N. Gras: *Sur les corps cubiques cycliques dont l'anneau des entiers est monotone.* Ann. Sci. Univ. Besançon, $3^e$ sér., Mathém., fasc. 6 (1973), 26pp.

M.-N. Gras: *Lien entre le groupe des unités et la monogénéité des corps cubiques cycliques.* Semaine. Théories des Nombres Besançon 1975-76. Publ. Math. Fac. Sci. Besançon, fasc. 1 (1976), 19pp.

E. Thomas: *Fundamental units for orders in certain cubic number fields.* J. reine u. angew. Math. 310 (1979), 33-55.

exceptional unit. So the $w$ in your blue case might be calledvery exceptional units! $\endgroup$ – Joe Silverman Oct 9 '18 at 16:36cardinalitiesof the two sets. (Too few characters for me to edit...) $\endgroup$ – GNiklasch Oct 9 '18 at 17:05