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.

Suppose there is some real number x that can't be written as the difference of transcendental numbers. Then for every transcendental number y there exists a unique algebraic number z = y+x. But this means there is an injective function f(a) = a+x from the transcendental numbers to the algebraic numbers, which implies the cardinality of the transcendental numbers is less than or equal to that of the algebraic numbers. But we know this is false because the cardinality of the algebraic numbers is strictly less than that of the transcendentals.

Q.E.D.