Answer
a) $39.3$
b) $171$
Work Step by Step
a) We know that air is a diatomic gas since it contains of $N_2, O_2,$ etc..
So we can consider it as an ideal gas.
For an ideal gas that undergoes an adiabatic process,
$$T_1V^{\gamma-1}_1=T_2V^{\gamma-1}_2$$
Hence,
$$\dfrac{T_1}{T_2}=\left[\dfrac{V_2}{V_1}\right]^{\gamma-1}$$
$$\left(\dfrac{T_1}{T_2}\right)^{\frac{1}{\gamma-1}}= \dfrac{V_2}{V_1} $$
Plugging the known;
$$\dfrac{V_2}{V_1}=\left(\dfrac{20+273}{1000+273}\right)^{\frac{1}{1.4-1}} =0.0254155 $$
This means that $V_2=0.025 V_1$ which means that $V_1$ is the maximum volume.
Therefore,
$$\dfrac{V_{\rm max}}{V_{\rm min}}=\dfrac{V_1}{V_2}=\dfrac{1}{0.0254155}$$
$$\dfrac{V_{\rm max}}{V_{\rm min}}=\color{red}{\bf3 9.3}$$
----
We also know, for an ideal gas that undergoes an adiabatic process, that
$$P_1V_1^\gamma=P_2V_2^\gamma$$
Hence,
$$\dfrac{P_2}{P_1}=\left[\dfrac{V_1}{V_2}\right]^\gamma$$
Plugging from above;
$$\dfrac{P_2}{P_1}=39.3^{1-4}=171$$
This means that $P_2=171P_1$ which means that $P_2$ is the maximum pressure.
Thus,
$$\dfrac{P_{\rm max}}{P_{\min}} =\color{red}{\bf171}$$
