We will show that there are exactly two solutions to this functional equation. First, take $x = 0$ to get
$$f(y) = f(0)g(1) + g(y),$$
so $g = f + C$ for some constant $C$. Since $f(1) = 3$, we deduce that
$$f(x+y) = (3+C)f(x) + f(y) + C, \quad x,y \in \mathbb{Q}. \qquad (15)$$
The equations $x + y + t = x + (y + t) = (x + t) + y$ imply that
$$f(x + y + t) = (3 + C)f(x) + f(y + t) + C = (3 + C)f(x + t) + f(y) + C.$$
Taking $y = x$, the last equation yields
$$(2 + C)f(x) = (2 + C)f(x + t), \quad x, t \in \mathbb{Q}.$$
It follows that either $f$ is constant (and equal to $3$) or $C = -2$. If $f \equiv 3$, then (15) implies that $3 = (3+C)4$, and so $C = -9/4$. This gives a constant solution, $(f,g) \equiv (3, 3/4)$, to our problem.

It remains to consider the case $C = -2$, and so (15) gives
$$f(x + y) = f(x) + f(y) - 2, \quad x, y \in \mathbb{Q}.$$
Writing $h := f - 2$, we get
$$h(x + y) = h(x) + h(y), \quad x, y \in \mathbb{Q}. \qquad (16)$$
If we take $y = 0$ in (16), we get $h(0) = 0$. If instead we take $x = -y$ in (16), we get $h(x) = -h(x)$. By repeatedly applying (16) to $h(mx)$, we get the identity $h(mx) = mh(x)$ for all $m \in \mathbb{N}$. By virtue of the identities $h(0) = 0$ and $h(-x) = x$, we can extend to $m \in \mathbb{Z}$:
$$h(mx) = mh(x), \quad x \in \mathbb{Q}, m \in \mathbb{Z}. \qquad (17)$$
Applying (17) with $mx/n$ in place of $x$ and $n$ in place of $m$, we see that $h(mx) = mh(mx/n)$ and so $mh(x) = mh(mx/n)$. We have shown that
$$h(rx) = rh(x), \quad x \in \mathbb{Q}, r \in \mathbb{Q}.$$
Rewriting this in terms of $f$ for $x = 1$, we get the identity $f(r) \equiv r + 2$, and so $g(r) \equiv r$. This is the second solution to the functional equation.