We apply Newton's second law to each block. Let's define the positive direction as $m_1$ accelerating up the incline and $m_2$ accelerating downward. The magnitude of acceleration, $a$, and the tension, $T$, are the same for both blocks.

For block $m_1$ on the incline, the net force along the incline is:

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

For the hanging block $m_2$, the net force is:

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

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

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

Solving for $a$ gives the acceleration:

$$a = \frac{g(m_2 - m_1 \sin\theta)}{m_1 + m_2}$$

The direction of motion depends on the sign of the numerator. If $m_2 > m_1 \sin\theta$, $a$ is positive and $m_2$ accelerates downward. If $m_2 < m_1 \sin\theta$, $a$ is negative and $m_2$ accelerates upward.

To find the tension, substitute the expression for $a$ back into equation (2):

$$T = m_2(g - a) = m_2\left(g - \frac{g(m_2 - m_1 \sin\theta)}{m_1 + m_2}\right)$$ $$T = m_2g \left(1 - \frac{m_2 - m_1 \sin\theta}{m_1 + m_2}\right) = m_2g \left(\frac{(m_1 + m_2) - (m_2 - m_1 \sin\theta)}{m_1 + m_2}\right)$$ $$T = \frac{m_1 m_2 g(1 + \sin\theta)}{m_1 + m_2}$$