Polynomials: gaining irreducibility by adding a constant

The case of degree $2$ polynomials in $\mathbb{R}$ or $\mathbb{Q}$ can be handled. Let $f(x) = ax^2 + bx$. By the quadratic formula, if $b^2 - 4ac < 0$, then we have $f(x) + c$ irreducible.

I have received important information from Michael Filaseta, with which we can answer:

1,2) Pick a prime $p$ big enough, put $c=p-f(0)$ and consider the polynomial $F(X)=f(X)+c$, which has $F(0)=p$. Specifically, if $f(X)=a_nX^n+\ldots+a_0$, by picking $p>|a_n|+\ldots+|a_1|$ we can guarantee that $F(X)$ has all its roots out of the unit circle, due to an iterative application of the reverse triangle inequality: $$|F(z)|\geq p-|a_1||z|-\ldots-|a_n||z^n|\geq p-(|a_1|+\ldots+|a_n|)>0,$$ where we have used that $|z|\leq1$.

Suppose $F$ factors as $F(X)=g(X)h(X)$; then $g(0)h(0)=p$ is a factorization of a prime, so for example $|g(0)|=1$. Therefore the absolute value of the product of the roots of $g$ is not greater than 1 (by Vieta, taking into account the leading coefficient of $g$). This implies that there is at least one root of $g$ inside the (closed) unit circle. But the roots of $g$ come from the roots of $F$, so we have reached a contradiction.

Now, as there are infinite primes bigger than $|a_n|+\ldots+|a_1|$, we know how to find an infinite number of $c$ such that $f+c$ is irreducible.

3) Hilbert's irreducibility theorem also answers 1), and gives the asymptotic behaviour: the polynomial $f+c$ is irreducible for almost every $c$. Concretely, if we denote $S(f,x):=\sum_{|c|\leq x, f+c\text{ irreducible}}1$ then we have $$S(f,x)=2x-o(x)$$ (the $2$ in $2x$ just comes from the fact that we consider $|c|\leq x$, so the density is computed with respect to $2x$).

In fact, it may be possible that using results close to Siegel's lemma one could prove $S(f,x)=2x-O(\sqrt{x}).$

5) For polynomials of degree 2 over the reals, as has already been mentioned, we can use the sign of the discriminant to guarantee the existence of infinite $c$, which asymptotically have $$\lim \frac{S(f,x)}{2x}=1/2,$$ as (if $a>0$, say) there is a $c_0$ such that if $c<c_0$ then $f+c$ factors, while if $c>c_0$ then $f+c$ is irreducible.

Now every real closed field is elementarily equivalent to the reals, and we can encode the condition on the discriminant in first order logic, so the same applies to real closed fields.