With any wiring job we have four considerations:-
The loop impedance both to earth and neutral.
The prospective short circuit current.
The volt drop.
The current carrying capacity of cable.
The first two are to ensure the protective device will trip and can handle the current when it does trip it is complicated compared with volt drop and in the main if volt drop is satisfied then so will the other two so I will deal with volt drop first.
For lighting under the new regulations we are allowed 3% and the regulations tell us the volt drop 1mmÂ² = 46mV/A/m, 1.5mmÂ² = 32mV/A/m, 2.5mmÂ² = 19mV/A/m and 4mmÂ² = 12mV/A/m so first job feet to meters 40 feet = 12.192 meters next is watts into amps this will depend on volts used 360 watt / volts = 1.56 amp at 230 volt or 30 amp at 12 volt so at 230volt 3% = 6.9 volt / 12.192 meters = 0.565945 volts per meter max / amps = 361.576mV/A/m which is higher than 1mmÂ² cable so smallest cable could be used. However if you do same calculations on 12 volt then we get 18.865mV/A/m so 4mmÂ² cable would need to be used. Since in real terms I expect all the lights would not be at cable end then 2.5mmÂ² cable would come within volt drop but not within current carrying capacity of cable. The current carrying capacity we get from tables and it depends on temp and route of cable for 230 volt system worst case is 8 amp for 1mmÂ² so well within it but for 12 volt we need at least 6mmÂ² and could need 10mmÂ² depending on ceiling and wall insulation. Most 12 volt systems would not accommodate cables that thick so separate cables would be needed for each lamp. So to conclude for 230volt 1mmÂ² cables for all lamps is ample and 12volt 1.5mmÂ² to each lamp min.
The first two considerations need to know earthing system and PSC plus ELI at origin if you want to know how I will go through it but it is unlikely to present a problem anyway.