The answer to the first question is positive. We may assume WLOG that $f$ is irreducible.
The Frobenius Density Theorem states that density of primes $p$ such that $f \mod p$ has a given decomposition (i.e. factors into irred. polynomials of prescribed degrees $n_1,\cdots,n_r$) is
$$\frac{\# \{ \sigma \in \text{Gal}(f) \mid \sigma \text{ has cycle structure }n_1,\cdots, n_r \} }{|\text{Gal}(f)|},$$

where $\text{Gal}(f)$ is the Galois group of $f$. See the discussion in this introductory paper by Sury.

By choosing the decomposition type of linear factors ($n_i = 1$), we see that for infinitely many primes $p$ (their density is $1$ over the size of the Galois group) we have that $f \mod p$ splits into linear factors (and in particular has zeroes). This is because the identity element of the Galois group of $f$ have cycle structure with $n_i=1$.

Remark: If you solely want the existence of infinitely many such primes (and not their density), The Frobenius density theorem may be avoided, see this discussion.