Inclined Massive Atwood Pulley (Frictionless)

Consider each block separately using Newton's second law.

Block on the incline ($m_1$):

The free body diagram for $m_1$ shows three forces:

Along the incline, the forces are $T_1$ up the incline and $m_1 g\sin\theta$ down the incline. Since the hanging mass accelerates downward, $m_1$ accelerates up the incline with magnitude $a$. Taking up the incline as positive,

Solving for $T_1$ gives

Hanging block ($m_2$):

The free body diagram for $m_2$ shows two forces:

The forces are $m_2 g$ downward and $T_2$ upward. Taking downward as positive (consistent with the given acceleration direction),

Solving for $T_2$ gives

Note that $T_1 \neq T_2$ because the pulley has mass and rotational inertia; the difference in tensions provides the net torque needed to angularly accelerate the pulley.