Apply Newton's second law to each block, defining the direction of motion as positive. Since $m_2 > m_1$, mass $m_2$ accelerates downward and $m_1$ accelerates upward with the same magnitude $a$.

For mass $m_1$:

$$T - m_1g = m_1a \quad (1)$$

For mass $m_2$:

$$m_2g - T = m_2a \quad (2)$$

To find the acceleration $a$, add equations (1) and (2) to eliminate the tension $T$:

$$(T - m_1g) + (m_2g - T) = m_1a + m_2a$$ $$m_2g - m_1g = (m_1 + m_2)a$$

Solving for $a$ gives:

$$a = \frac{m_2 - m_1}{m_1 + m_2}g$$

To find the tension $T$, substitute the expression for $a$ back into equation (1):

$$T = m_1a + m_1g = m_1(a + g)$$ $$T = m_1\left(\frac{m_2 - m_1}{m_1 + m_2}g + g\right)$$ $$T = m_1g\left(\frac{m_2 - m_1}{m_1 + m_2} + 1\right) = m_1g\left(\frac{m_2 - m_1 + m_1 + m_2}{m_1 + m_2}\right)$$ $$T = m_1g\left(\frac{2m_2}{m_1 + m_2}\right) = \frac{2m_1m_2g}{m_1 + m_2}$$