Answer:
1) Volume occupied by the spheres are equal therefore the three tanks contains the same volume of water
2) 
Step-by-step explanation:
1) Here we have;
First tank A
Volume of tank = x³
The volume of the sphere = 
However, the diameter of the sphere = x therefore;
r = x/2 and the volume of the sphere is thus;
volume of the sphere =
= 
For tank B
Volume of tank = x³
The volume of the spheres = 
However, the diameter of the spheres 2·D = x therefore;
r = x/4 and the volume of the sphere is thus;
volume of the spheres = 
For tank C
Volume of tank = x³
The volume of the spheres = 
However, the diameter of the spheres 4·D = x therefore;
r = x/8 and the volume of the sphere is thus;
volume of the spheres = 
Volume occupied by the spheres are equal therefore the three tanks contains the same volume of water
2) For the 4th tank, we have;
number of spheres on side of the tank, n is given thus;
n³ = 512
∴ n = ∛512 = 8
Hence we have;
Volume of tank = x³
The volume of the spheres = 
However, the diameter of the spheres 8·D = x therefore;
r = x/16 and the volume of the sphere is thus;
volume of the spheres = 
Amount of water remaining in the tank is given by the following expression;
Amount of water remaining in the tank = Volume of tank - volume of spheres
Amount of water remaining in the tank = 
.