Gelfond-Schneider theorem
If $a,b$ are algebraic; $a \ne 0,1$; $b$ irrational then (any value of) $a^b$ is transcendental.
Let $A$ be algebraic irrational.
We claim that $\cos(A\pi)$ transcendental. Assume not:
$\cos(A\pi)$ is algebraic.
Then $\sin(A\pi) = \pm\sqrt{1-\cos^2(A\pi)}$ is algebraic. Then $e^{iA\pi}=\cos(A\pi)+i\sin(A\pi)$ is algebraic.
Also $e^{iA\pi} \ne 0$ and $e^{iA\pi} \ne 1$.
Now $2/A$ is algebraic irrational.
Note
$$
\big(e^{iA\pi}\big)^{(2/A)} = e^{2\pi i} = 1
$$
is algebraic. This contradicts Gelfond-Schneider.
It’s not precisely easy, but here is a piecewise linear order-preserving bijection.
First, suppose that $p,q,r,s\in\Bbb Q$, $p<q$, and $r<s$. Let
$$h_{p,q,r,s}:(p,q)\to(r,s)\setminus\Bbb Q:x\mapsto r+\frac{s-r}{q-p}(x-p)\,$$
$h_{p,q,r,s}$ is an order-preserving bijection that takes rationals to rationals and irrationals to irrationals, so its restriction to $(p,q)\setminus\Bbb Q$ is a bijection to $(r,s)\setminus\Bbb Q$.
Now let $a_n=-2^{-n}$ for $n\in\Bbb N$, and let $\langle b_n:n\in\Bbb N\rangle$ be a strictly increasing sequence in $\Bbb Q$ converging to $\pi$ such that $b_0=-1=a_0$. Let $f_0$ be the identity map on $(\leftarrow,-1)\setminus\Bbb Q$, and for $n\in\Bbb Z^+$ let $f_n=h_{a_{n-1},a_n,b_{n-1},b_n}$. Then
$$f=\bigcup_{n\in\Bbb N}f_n:(\leftarrow,0)\setminus\Bbb Q\to(\leftarrow,\pi)\setminus\Bbb Q$$
is an order-preserving bijection.
Best Answer
Here’s an answer to part of (2).
The order type of $\mathbb{P}$, the irrationals, is homogeneous, because $\langle\mathbb{P},\le\rangle$ is order-isomorphic to $\langle\mathbb{Z}^\omega,\preceq\rangle$, where $\preceq$ is the lexicographic order, which is order-homogeneous.
To construct the order-isomorphism, enumerate the rationals as $\mathbb{Q}=\{q_n:n\in\omega\}$. Recursively construct open intervals $I_\sigma$ for $\sigma\in\mathbb{Z}^{<\omega}$ to satisfy the following conditions.
Clauses (4)-(6) ensure that for each $\sigma\in\mathbb{Z}^{<\omega}$ and $n\in\mathbb{Z}$, $\operatorname{cl}I_{\sigma^\frown n}\subseteq I_\sigma$, so for each $\sigma\in\mathbb{Z}^\omega$, $$\bigcap\limits_{n\in\omega}I_{\sigma\upharpoonright n}\ne\varnothing\;.\tag{1}$$ Clause (7) ensures that the intersection in $(1)$ is at most a singleton, so for each $\sigma\in\mathbb{Z}^\omega$ there is a unique $h(\sigma)\in\mathbb{R}$ such that $$\bigcap\limits_{n\in\omega}I_{\sigma\upharpoonright n}=\{h(\sigma)\}\;.$$ Finally, clause (3) ensures that $h(\sigma)\in\mathbb{P}$, so $h:\mathbb{Z}^\omega\to\mathbb{P}$.
To see that $h$ is an injection, suppose that $\sigma,\tau\in\mathbb{Z}^\omega$, and $\sigma\ne\tau$. Let $n\in\omega$ be minimal such that $\sigma\upharpoonright n\ne\tau\upharpoonright n$; then by construction $I_{\sigma\upharpoonright n}\cap I_{\tau\upharpoonright n}=\varnothing$, so $h(\sigma)\ne h(\tau)$. To see that $h$ is surjective, simply observe that for any $x\in\mathbb{P}$ and any $n\in\omega$ there is a unique $\sigma\in\mathbb{Z}^n$ such that $x\in I_\sigma$. Thus, $h$ is a bijection, and it only remains to check that $h$ is order-preserving.
Suppose that $\sigma,\tau\in\mathbb{Z}^\omega$ with $\sigma\prec\tau$. Let $n\in\omega$ be minimal such that $\sigma(n)\ne\tau(n)$, and let $\varphi=\sigma\upharpoonright n=\tau\upharpoonright n$. Then $h(\sigma),h(\tau)\in I_\varphi$, $h(\sigma)\in I_{\varphi^\frown \sigma(n)}$, $h(\tau)\in I_{\varphi^\frown \tau(n)}$, and $\sigma(n)<\tau(n)$, so $h(\sigma)<h(\tau)$ by clause (5).
Finally, to see that $\langle\mathbb{Z}^\omega,\preceq\rangle$ is order-homogeneous, let $\sigma,\tau\in\mathbb{Z}^\omega$, and define $\delta\in\mathbb{Z}^\omega$ by $\delta(n)=\tau(n)-\sigma(n)$. Then the shift $$s:\mathbb{Z}^\omega\to\mathbb{Z}^\omega:\varphi\mapsto\langle\varphi(n)+\delta(n):n\in\omega\rangle$$ is an order-automorphism of $\langle\mathbb{Z}^\omega,\preceq\rangle$ taking $\sigma$ to $\tau$.