(2) ∵ L 1: L2 = 3: 1，F 1=2.4N，F 1L 1=GL2。
∴g=f 1l 1l2=2.4n×3 1=7.2n，
Ore quality m = gg = 7.2n10n/kg = 0.72kg;
(3)A, putting the ore into a beaker, adding a proper amount of water to immerse the ore, and marking the position where the water surface reaches;
B. taking out the ore;
C. Add water to the beaker with a measuring cylinder until the water surface reaches the marked position.
(4) As can be seen from Figure 3, the bottom of the liquid concave surface in the measuring cylinder is flush with the 36mL scale line, so the volume of the ceramic tile is100ml+50ml-36ml =114ml =14cm3.
Ore density: ρ = MV = 720g114cm3 = 6.32g/cm3 ≈ 6.32x103kg/m3.
So the answer is: (1) The mass of ore exceeds the range of the balance; (2)0.72; (3) the water surface reaches the marked position; (3)6.32× 103.