Answer
a) $2034\;\rm L$
b) $85\;\rm min$
c) $26\;\rm min$
Work Step by Step
a) We can assume that air is an ideal gas.
The ideal gas law is given by
$$PV=nRT$$
The author here asks about the change of air volume when we change pressure while the temperature remains constant.
So,
$$P_1V_1=P_2 V_2$$
Solving for $V_2$;
$$V_2=\dfrac{P_1V_1}{P_2 }$$
Plugging the given;
$$V_2=\dfrac{180\cdot 11.3}{1}=\boxed{\bf 2034\;\rm L}$$
b) The volume of air consumes in one minute
$$V_{min}=2 \cdot 12 =24\;\rm L$$
The time he takes him to finish the tank is
$$\Delta t=\dfrac{V_2}{V_{min}}=\dfrac{2034}{24}$$
$$\Delta t\approx \boxed{\bf 85\;\rm min}$$
c) We need to find the volume of air in this new situation.
The tank is 23 m below the water surface which means that the pressure increased and the temperature decreased to 10$^\circ\;\rm C$.
Using the same ideal gas law we used above, first, we need to find the new pressure.
$$P_2=P_a+h\rho_w g\tag 1$$
whereas $P_a$ is atmospheric pressure and $\rho_w$ is the density of the sea water.
From the ideal gas law;
$$\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}$$
Solving for $V_2$;
$$V_2=\dfrac{P_1V_1T_2}{P_2T_1} $$
Plugging $P_2$ from (1) and then plugging the given and recall to convert pressures and temperatures to SI units.
$$V_2=\dfrac{180\cdot 1.013\times10^5\cdot11.3\cdot 291 }{\left([1.013\times10^5]+[23\cdot 1030\cdot 9.8]\right)\cdot 283} =\;\rm \bf 635.4\;L$$
The time he takes him to finish the tank is
$$\Delta t=\dfrac{V_2}{V_{min}}=\dfrac{635.4}{24}$$
$$\Delta t\approx \boxed{\bf 26\;\rm min}$$