$\mathbb{Q}(\sqrt{2})$ carries exactly two distinct field orders, exchanged by the conjugation $\sqrt{2} \mapsto -\sqrt{2}$

$u \in \mathbb{R}$ satisfies $u > 0$ and $u^2 = 2$, and $u \notin \mathbb{Q}$.

In every ordered field the positivity of a rational is forced: $n \cdot 1 > 0$ for $n \ge 1$, so for positive integers $p, q$ the element $(p\cdot 1)(q\cdot 1)^{-1}$ is positive, and hence a rational is in the positive cone exactly when it is positive in the usual sense.

$F$ is a subfield of $\mathbb{R}$: it contains $0$ and $1$, is closed under subtraction, and $(a+bu)(c+du) = (ac + 2bd) + (ad + bc)u$ gives closure under multiplication; for $a + bu \ne 0$ one has $a^2 - 2b^2 \ne 0$, since $b \ne 0$ would otherwise give $(a/b)^2 = 2$ against [L2] while $b = 0$ forces $a \ne 0$, and then $(a+bu)^{-1} = (a - bu)(a^2-2b^2)^{-1} \in F$.

The representation is unique: $a + bu = a' + b'u$ with $b \ne b'$ would give $u = (a-a')(b'-b)^{-1} \in \mathbb{Q}$, against step 1.1; so $b = b'$ and then $a = a'$.

$\sigma$ is therefore a well-defined map $F \to F$, and it is a field automorphism: it is additive by inspection, $\sigma(1) = 1$, and $\sigma(a+bu)\sigma(c+du) = (ac+2bd) - (ad+bc)u = \sigma\big((a+bu)(c+du)\big)$; moreover $\sigma \circ \sigma$ is the identity, so $\sigma$ is a bijection.

Let $Q$ be any positive cone on $F$. Since $u \ne 0$, exactly one of $u \in Q$, $-u \in Q$ holds.

$Q$ is determined by that choice. Suppose $u \in Q$ (the other case is the same with $u$ replaced by $-u$, which also squares to $2$). Let $x = a + bu \ne 0$. If $b = 0$ then $x$ is a nonzero rational and step 1.2 decides it. If $b \ne 0$ then $x = b(u + c)$ with $c := a/b \in \mathbb{Q}$, and by [L4] the membership of $x$ is decided by those of $b$ and of $u + c$; for $c \ge 0$ one has $u + c \in Q$, while for $c < 0$, writing $e := -c > 0$, the identity $(u-e)(u+e) = 2 - e^2$ with $u + e \in Q$ and $2 - e^2 \ne 0$ shows that $u - e \in Q$ exactly when $2 - e^2 > 0$, a condition on a rational decided by step 1.2. So $Q$ is uniquely determined, and there are at most two positive cones on $F$.

Both occur. $P_1$ is a positive cone on $F$, being the restriction to the subfield $F$ of the positive cone of $\mathbb{R}$; and $P_2 = \sigma^{-1}(P_1)$ is one because $\sigma$ is a field automorphism, so trichotomy and closure transfer along it. They are distinct: $u \in P_1$ by step 1.1, whereas $\sigma(u) = -u \notin P_1$, so $u \notin P_2$.

Hence $F$ carries exactly two positive cones, $P_1$ and $P_2$, and since $\sigma$ is an involution, $P_2 = \sigma(P_1)$ and $P_1 = \sigma(P_2)$: the conjugation exchanges the two orders.