Every Real number is expressible in terms of differences of two transcendentals

If $x$ is algebraic, then take $\alpha=x\pi$ and $\beta=x(\pi-1)$.

If $x$ is transcedental, then take $\alpha=2x$ and $\beta=x$.

Yes, since the set of algebraic numbers is countable.

Let $\mathbb{T}$ denote the set of transcendental numbers, and for $\alpha\in \mathbb{T}$ let $f(\alpha)=\alpha-x$. Then $f$ is an injection from $\mathbb{T}$ to $\mathbb{R}$, so its range $ran(f)$ is uncountable since $\mathbb{T}$ is. But since $\mathbb{R}\setminus\mathbb{T}$ (the set of algebraic numbers) is countable, this means $ran(f)$ contains some element of $\mathbb{T}$.

So let $\alpha\in\mathbb{T}$ be such that $f(\alpha)\in\mathbb{T}$, and set $\beta=f(\alpha)$; then $\alpha, \beta$ are transcendental and $\alpha-\beta=x$.

Once we know that $e$ is transcendental, we also know that $e^2$ and $e^2-e$ are transcendental because given a polynomial satisfied by either one we could find a polynomial for $e$. Then given algebraic $a \in \Bbb R$ we can write $a = (a+e)-e$. Given transcendental $a \in \Bbb R$ we can either write $a=(a+e)-e$ or $a=(a+e^2)-e^2$. We know that at least one of $a+e, a+e^2$ is transcendental because if they were both algebraic, so would $e^2-e$ be.