We apply Newton's second law by analyzing the horizontal and vertical components of the forces. Let the acceleration be $a$, the normal force be $N$, and gravitational acceleration be $g$.

[Q1] The only horizontal force is the x-component of the applied force $F$.

$$ \sum F_x = F \cos\theta = ma $$

Solving for the acceleration $a$:

$$ a = \frac{F \cos\theta}{m} $$

[Q2] The block lifts off the floor when the normal force $N$ becomes zero. The vertical forces are the upward normal force $N$, the upward vertical component of $F$, and the downward force of gravity $mg$.

$$ \sum F_y = N + F \sin\theta - mg = 0 $$

Setting $N=0$ for liftoff:

$$ F \sin\theta - mg = 0 $$

Solving for the liftoff force $F$:

$$ F = \frac{mg}{\sin\theta} $$

[Q3] To find the acceleration just before liftoff, we substitute the liftoff force from [Q2] into the acceleration equation from [Q1].

$$ a = \frac{1}{m} \left( \frac{mg}{\sin\theta} \right) \cos\theta $$

Simplifying the expression gives the acceleration at liftoff:

$$ a = g \frac{\cos\theta}{\sin\theta} = g \cot\theta $$