Here's the full problem that I'm trying to solve, and my partial progress:
Let $f(X) = X^5+4X^3+9X+3 \in \mathbb{Q}[X]$, and let $f_{2}(X) \in \mathbb{F}_{2}[X]$ and $f_{3}(X) \in \mathbb{F}_{3}[X]$ be reductions of $f$ over $\mathbb{F}_{2}$ and $\mathbb{F}_{3}$. Let $K$ be the splitting field of $f$ over $\mathbb{Q}$, $K_{2}$ be the splitting field of $f_{2}$ over $\mathbb{F}_{2}$ and $K_{3}$ bt the splitting field of $f_{3}$ over $\mathbb{F}_{3}$. Find $|K_{2}|$, $|K_{3}|$ and $\text{Gal}(K/\mathbb{Q})$.
By checking that $(f,f')=1$, I find that $f$ is separable, so $K$ is the splitting field of a separable polynomial over $\mathbb{Q}$, therefore $K/\mathbb{Q}$ is a Galois extension.
Furthermore, $f_{2}(X) = X^5+X+1$ and $f_{3}(X) = X^5+X^3$. By a straightforward check, I find that $f_{2}$ has no zeros in $\mathbb{F}_{2}$, and that it can't be factored as $3+2$, so I find that it's irreducible. Therefore, by a known theorem, $\text{Gal}(K/\mathbb{Q})$ contains a $5$-cycle. Also, I find that if $\zeta$ is a zero of $f_{2}$, then $[\mathbb{F}_{2}[\zeta] : \mathbb{F}_{2} ] = 5$, which means $K_{2}$ has at least $32$ elements.
Now, $f_{3}(X) = X^3(X^2+1)$, so $K_{3} = \mathbb{F}_{3}[i]$, and $|K_{3}|=9$. Of course, $i$ here is not the "classic" $i$, just a zero of $X^2+1$, which is irreducible in $\mathbb{F}_{3}$.
Also, I'm not sure how much help this is, but $f'(X) = 5X^4+12X^2+9 > 0$, so $f$ has only $1$ real zero and $2$ pairs of complex conjugates, so the restriction of conjugation in $\mathbb{C}$ to $K$ is not $Id$, but a product of two transpositions, so $\text{Gal}(K/\mathbb{Q})$ also contains a product of $2$ transpositions.
Is there a way to determine $|K_{2}|$ and $\text{Gal}(K/\mathbb{Q})$ from here?
For the case when the polynomial is over $\mathbb{Q}$, as you have noticed we have a complex pair as roots. However as there are $4$ complex roots this isn't enough to determine whether the Galois group is $D_5$, $\mathbb{Z}_5 \rtimes \mathbb{Z}_4$, $A_5$ or $S_5$. The easiest way to go here is to compute the discriminant of the polynomial. In general calculating the discriminant of a quintic is a tedious job, but here we are lucky as couple of the coefficients are $0$. In general for polynomials of the form $x^5 + bx^3 + dx + e$ the discriminant is given by:
$$D =108 b^5 e^2+16 b^4 d^3-900 b^3 d e^2-128 b^2 d^4+2000 b d^2 e^2+256 d^5+3125 e^4$$
A lengthy calculation yields $D = 7080453 = 3^4 \cdot 61 \cdot 1433$, which is a non-square. This means that $\sqrt{D} \in K$ isn't fixed by all automorphisms. However elements from $A_5$ fix it, so we conclude that $K \not = A_5, D_5$, as $D_5 \le A_5$.
To discard $\mathbb{Z}_5 \rtimes \mathbb{Z}_4$ you can use the Cayley's resolvent, as mentioned here to prove that the quintic isn't solvable in radicals, thus its Galois group can't be $\mathbb{Z}_5 \rtimes \mathbb{Z}_4$, a solvable group. Thus we conclude that $\operatorname{Gal}(K/\mathbb{Q}) \cong S_5$.
Another method, which is easier, but more advanced is to use Dedekind's Theorem. Factoring $f$ modulo $37$ we get that:
$$x^5 + 4x^3 + 9x + 3 = (x+25)(x+28)(x+29)(x^2+29x+14) \in \mathbb{F}_{37}[x]$$
Thus we conclude that $\operatorname{Gal}(K/\mathbb{Q})$, viewed as a subgroup of $S_5$ contains a transposition. On the other hand it's well-known (mentioned in the link above too) that $S_p$ is generated by a $p$-cycle and transposition for prime $p$. Thus we conclude again that $\operatorname{Gal}(K/\mathbb{Q}) \cong S_5$.