Let $a$ be the magnitude of the system's acceleration. All surfaces are frictionless.

[Q1] Assume the system accelerates with $m_1$ moving to the right and $m_2$ moving down the incline. The leftward force $F$ opposes this motion for $m_1$. Applying Newton's second law to each mass in the direction of motion:

$$T - F = m_1 a \quad \quad (1)$$ $$m_2 g \sin\theta - T = m_2 a \quad \quad (2)$$

To find the tension $T$, we first eliminate the acceleration $a$. From (1), $a = (T-F)/m_1$. Substituting this into (2):

$$m_2 g \sin\theta - T = m_2 \left(\frac{T-F}{m_1}\right)$$

Multiplying by $m_1$ and rearranging to solve for $T$:

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

[Q2] The cord will become slack if block $m_1$ is pulled to the right with an acceleration that block $m_2$ cannot match. The limiting condition occurs when the acceleration of $m_1$ under the force $F_{max}$ (if the cord were cut) equals the natural free-fall acceleration of $m_2$ down the incline.

The unconstrained acceleration of $m_1$ due to a rightward force $F_{max}$ is:

$$a_1 = \frac{F_{max}}{m_1}$$

The unconstrained acceleration of $m_2$ down the frictionless incline is:

$$a_2 = g \sin\theta$$

Setting these accelerations equal gives the maximum force before the cord would become slack:

$$\frac{F_{max}}{m_1} = g \sin\theta$$ $$F_{max} = m_1 g \sin\theta$$