Every Real number is expressible in terms of differences of two transcendentals Is it true that for every real number $x$ there exist transcendental numbers $\alpha$ and $\beta$ such that $x=\alpha-\beta$?
(it is true if $x$ is an algebraic number). 

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 $\alpha-\beta=x$.
