Let $v_0$ be the initial upward speed, $v_f = \frac{3}{4}v_0$ be the final speed upon return, $m$ be the mass, $f$ be the constant air resistance, and $h$ be the maximum height.

[Q1] During upward motion, the net downward force is $mg+f$. The upward acceleration is $a_{up} = -(g + f/m)$. Using the kinematic equation $v^2 = u^2 + 2as$ for the upward trip:

$$0^2 = v_0^2 + 2a_{up}h \implies v_0^2 = 2(g + f/m)h$$

During downward motion, the net downward force is $mg-f$. The downward acceleration is $a_{down} = g - f/m$. Using the same kinematic equation for the downward trip:

$$v_f^2 = 0^2 + 2a_{down}h \implies v_f^2 = 2(g - f/m)h$$

Dividing the second equation by the first:

$$\frac{v_f^2}{v_0^2} = \frac{2(g - f/m)h}{2(g + f/m)h} = \frac{g - f/m}{g + f/m}$$

Substitute the given velocity ratio, $v_f/v_0 = 3/4$:

$$(\frac{3}{4})^2 = \frac{9}{16} = \frac{g - f/m}{g + f/m}$$

Solving for $f/m$:

$$9(g + f/m) = 16(g - f/m) \implies 9g + 9f/m = 16g - 16f/m$$ $$25f/m = 7g \implies f = \frac{7}{25}mg$$

The ratio of air resistance to gravity is:

$$\frac{f}{mg} = \frac{7}{25}$$

[Q2] The maximum height with air resistance is $h = \frac{v_0^2}{2(g + f/m)}$. In a vacuum ($f=0$), the maximum height is $h_0 = \frac{v_0^2}{2g}$. The ratio of the heights is:

$$\frac{h}{h_0} = \frac{v_0^2 / [2(g + f/m)]}{v_0^2 / (2g)} = \frac{g}{g + f/m}$$

Substitute the result from Q1, $f/m = (7/25)g$:

$$\frac{h}{h_0} = \frac{g}{g + (7/25)g} = \frac{1}{1 + 7/25} = \frac{1}{32/25} = \frac{25}{32}$$