4 How much can I save on cabling using DC?

With the same insulation requirements and the same power, about 50% copper cross-section can be saved. 

Example for a 7.5 kW three-phase motor connected with a frequency converter (cos⁡φ=0.7,〖 η〗_Motor=0.887,η_Inverter=0.93): 

AC:
    400 V AC, 4 conductors (L1, L2, L3, PE)
    Mains Current = 20 A (Lenze Manual for i550 Inverter) → this requires 2.5 mm² Cu-cross-section
    4 × 2.5 mm² = 10 mm² Cu-cross-section

DC:
    600 V, 3 conductors (Plus, Minus, PE)
    15.2 A (P=U×I×η ) require 1.5 mm² Cu-cross-section
    3 × 1.5 mm² = 4.5 mm² Cu-cross-section

55% less copper!