Apply Newton's second law to each mass and the rotational equivalent to the pulley. Let $a$ be the linear acceleration of the masses, and $T_1$ and $T_2$ be the tensions on the sides of $m_1$ and $m_2$ respectively. The pulley is a solid disk with moment of inertia $I = \frac{1}{2}mR^2$.

For mass $m_1$ (moving down):

$$m_1g - T_1 = m_1a$$

For mass $m_2$ (moving up):

$$T_2 - m_2g = m_2a$$

For the pulley's rotation, the net torque is $\tau = (T_1 - T_2)R$. Using $\tau = I\alpha$ and the no-slip condition $a = \alpha R$:

$$(T_1 - T_2)R = I\alpha = \left(\frac{1}{2}mR^2\right)\left(\frac{a}{R}\right)$$ $$T_1 - T_2 = \frac{1}{2}ma$$

Solving the three linear equations for $a, T_1,$ and $T_2$: Subtract the equation for $m_2$ from the equation for $m_1$:

$$(m_1g - T_1) - (T_2 - m_2g) = m_1a - m_2a$$ $$(m_1 + m_2)g - (T_1 + T_2) \rightarrow \text{This is not helpful.}$$

Instead, substitute for $T_1$ and $T_2$ into the torque equation. From the first two equations, $T_1 = m_1(g-a)$ and $T_2 = m_2(g+a)$.

$$m_1(g-a) - m_2(g+a) = \frac{1}{2}ma$$ $$(m_1 - m_2)g = \left(m_1 + m_2 + \frac{1}{2}m\right)a$$ $$a = \frac{(m_1 - m_2)g}{m_1 + m_2 + \frac{1}{2}m}$$

Substitute $a$ back to find the tensions:

$$T_1 = m_1(g-a) = m_1g\left(1 - \frac{m_1 - m_2}{m_1 + m_2 + \frac{1}{2}m}\right) = \frac{m_1g(2m_2 + \frac{1}{2}m)}{m_1 + m_2 + \frac{1}{2}m}$$ $$T_2 = m_2(g+a) = m_2g\left(1 + \frac{m_1 - m_2}{m_1 + m_2 + \frac{1}{2}m}\right) = \frac{m_2g(2m_1 + \frac{1}{2}m)}{m_1 + m_2 + \frac{1}{2}m}$$