The initial momentum of $A$ is $p_a = 48 \times 7 = 336\,\text{kgms}^{-1}$, and the initial momentum of $B$ is zero since it is not moving.
Since momentum is always conserved, the total initial momentum, $336\,\text{kgms}^{-1}$, must equal the total final momentum, $1.4m_a + 1.4m_b$.
I.e
$$336 = 1.4m_a + 1.4m_b$$
So
$$m_b = \frac{336 - 1.4 \times 48}{1.4} = 192\,\text{kg}$$
as you found.
We can now calculate the impulse of $A$ and $B$.
$$\text{Imp}_a = \Delta p = 1.4\times48 - 336 = -268.8\,\text{kgms}^{-1}$$
$$\text{Imp}_b = \Delta p = 1.4\times192 - 0 = 268.8\,\text{kgms}^{-1}$$
As you can see $\text{Imp}_b = -\text{Imp}_a$.
This can be explained by Newton's Third Law: the force exerted on $B$ by $A$ is equal and opposite to the force exerted on $A$ by $B$ during the collision. And since they collide for the same time the impuse, $\int Fdt$, has the same magnitude for both objects, but different directions (signs).
Getting to your specific questions...
Is the impulse equal to $0$ since the change of momentum doesn't change in an inelastic collision?
Remember their is only a particular object's impulse, just like you can't have a force without an object. Although the total momentum of the system does not change, since it is conserved, clearly the momentum of the individual objects changes. This means an impulse has been delivered to both of them.
Is it possible not to have an impulse during a collision?
If you define a collision as objects changing speeds, then momentum must have changed so impulses must have been delivered in every collision.