Since the objects are connected and move together, they share a common acceleration, $a$. We apply Newton's second law, $\Sigma F=ma$, to two different subsystems.

Consider the subsystem of masses $m_1$ and $m_2$. The net force is the tension $T_2$.

$$T_2 = (m_1 + m_2)a$$

Consider the subsystem of masses $m_3$ and $m_4$. The net force is the difference between the forward tension $T_4$ and the backward tension $T_2$.

$$T_4 - T_2 = (m_3 + m_4)a$$

To eliminate the acceleration $a$, we can solve for $a$ in each equation and set them equal, or simply divide the two equations:

$$\frac{m_1 + m_2}{m_3 + m_4} = \frac{T_2}{T_4 - T_2}$$

Now, we solve for the unknown mass $m_2$:

$$m_1 + m_2 = (m_3 + m_4) \frac{T_2}{T_4 - T_2}$$ $$m_2 = (m_3 + m_4) \frac{T_2}{T_4 - T_2} - m_1$$