Let $V$ be a geometrically integral variety over a field $K$.
I consider the following properties:

(1) There exists a dominant rational map $\mathbb{P}_K^n\dashrightarrow V$ for $n={\rm dim}(V)$.

(2) There exists a dominant rational map $\mathbb{P}_K^n\dashrightarrow V$ for some $n$.  ($V$ is unirational)

(3) There exists a dominant rational map $\mathbb{P}_K^n\dashrightarrow V$ with geometrically
integral generic fiber, for some $n$.

(4) There exists a dominant rational map $\mathbb{P}_K^n\dashrightarrow V$ with a right inverse $V\dashrightarrow\mathbb{P}_K^n$, for some $n$. ($V$ is retract rational)

(5) There exists a birational map $\mathbb{P}_K^n\dashrightarrow V\times\mathbb{P}_K^m$ for some $m,n$. ($V$ is stably rational)

(6) There exists a birational map $\mathbb{P}_K^n\dashrightarrow V$ for some $n$. ($V$ is rational)

We have that $(6)\Rightarrow(5)\Rightarrow(4)\Rightarrow(3)\Rightarrow(2)\Leftrightarrow(1)$. For curves all of these properties are equivalent, but they diverge in higher dimension. From browsing the literature I gather that it is known that $(2)\not\Rightarrow(4)$ and $(5)\not\Rightarrow(6)$, and it is expected that $(4)\not\Rightarrow(5)$.

However, I am interested in property (3), which I could not find anywhere in the literature.

**Q1:** Does property (3) occur in the literature? Does it have a name? Is it equivalent to either (2) or (4)?

More concretely, I wonder whether every real surface that satisfies (3) also satisfies (5), or even (6), or whether one should not expect this (e.g. because one expects $(4)\Rightarrow(5)$ to fail already for real surfaces). To slightly simplify and phrase things a bit more down to earth:

**Q2:** Is every intermediate field $F$ of $\mathbb{R}(X,Y,Z)/\mathbb{R}$ which is algebraically closed in $\mathbb{R}(X,Y,Z)$ purely transcendental over $\mathbb{R}$?

Of course this is clear for $F$ of transcendence degree $0$, $1$ or $3$ over $\mathbb{R}$, so it is really just a question about surfaces. The equivalent question over $\mathbb{C}$ has a positive answer, as every unirational complex surface is known to be rational. The closest to a counterexample I found in the literature is the surface over $\mathbb{R}$ given by $x^2+y^2=f(z)$ with $f$ of degree $3$ with three real roots, which I think satisfies (2) but not (5), but I don't know if it satisfies (3).