How many of the cubes have exactly 4 face painted?

Let us see the changes due to removal of cube from corner-
Number of vertices with three faces exposed (Painted) is 7 + 3 = 10
Number of Cubes with 2 sides exposed (Painted): In general one edge gives us 4 (n - 2 in general case) cubes with two face painted but in this case out of 12 edges only 9 edges will give us 4 cubes in one edge and remaining 3 edges will give us 3 cubes from one edge, hence total number of edge is 9 x 4 + 3 x 3 = 45
Number of Cubes with 1 side exposed (Painted): It will remain same as normal case i.e. 6(4²) = 96
Number of Cubes with no sides exposed (Painted) is 4³ = 64
From the above observation:
From the above explanation number of the cubes with at least 2 faces painted is 45 + 10 = 55.

Let us see the changes due to removal of cube from corner-
Number of vertices with three faces exposed (Painted) is 7 + 3 = 10
Number of Cubes with 2 sides exposed (Painted): In general one edge gives us 4 (n - 2 in general case) cubes with two face painted but in this case out of 12 edges only 9 edges will give us 4 cubes in one edge and remaining 3 edges will give us 3 cubes from one edge, hence total number of edge is 9 x 4 + 3 x 3 = 45
Number of Cubes with 1 side exposed (Painted): It will remain same as normal case i.e. 6(4²) = 96
Number of Cubes with no sides exposed (Painted) is 4³ = 64
From the above observation:
From the above explanation number of the cubes with at most 2 faces painted is 64 + 96 + 45 = 205.
Or else 215 - 10 = 205

Let us see the changes due to removal of cube from corner-
Number of vertices with three faces exposed (Painted) is 7 + 3 = 10
Number of Cubes with 2 sides exposed (Painted): In general one edge gives us 4 (n - 2 in general case) cubes with two face painted but in this case out of 12 edges only 9 edges will give us 4 cubes in one edge and remaining 3 edges will give us 3 cubes from one edge, hence total number of edge is 9 x 4 + 3 x 3 = 45
Number of Cubes with 1 side exposed (Painted): It will remain same as normal case i.e. 6(4²) = 96
Number of Cubes with no sides exposed (Painted) is 4³ = 64
From the above observation:
From the above explanation number of the cubes with 2 faces painted is 45.

Let us see the changes due to removal of cube from corner-
Number of vertices with three faces exposed (Painted) is 7 + 3 = 10
Number of Cubes with 2 sides exposed (Painted): In general one edge gives us 4 (n - 2 in general case) cubes with two face painted but in this case out of 12 edges only 9 edges will give us 4 cubes in one edge and remaining 3 edges will give us 3 cubes from one edge, hence total number of edge is 9 x 4 + 3 x 3 = 45
Number of Cubes with 1 side exposed (Painted): It will remain same as normal case i.e. 6(4²) = 96
Number of Cubes with no sides exposed (Painted) is 4³ = 64
From the above observation:
From the above explanation number of the cubes with 0 faces painted is 64.

Out of 6 faces of 5 faces are exposed and those were painted.
Number of vertices with three faces exposed (Painted) is 4
Number of vertices with 2 faces exposed (Painted) is 4
Number of vertices with 1 faces exposed (Painted) is 0
Number of vertices with 0 faces exposed (Painted) is 0
Number of sides with 2 sides exposed (Painted) is 8
Number of sides with 1 sides exposed (Painted) is 4
Number of sides with no sides exposed (Painted) is 0
From the above observation:
Number of cubes with 3 faces Painted is 4
Number of cubes with 2 faces Painted is given by sides which is exposed from two sides, out of 8 such edges 4 vertical edges will give us 6 cubes per edge and 4 edges from top surface will give us 5 such cubes from each edge and required number of cubes is 6 x 4 + 4 x 5 = 44.
Number of cubes with 1 face Painted is given by faces which is exposed from one sides four vertical faces will give us 6 x 5 = 30 cubes per face and top face will give us 5 x 5 = 25 and required number of cubes is 30 x 4 + 25 x 1 = 145
Number of cubes with 0 face Painted is given by difference between total number of cubes - number of cubes with at least 1 face painted = 343 - 4 - 44 - 145 = 150
In other words number of cubes with 0 painted is 6 x 5 x 5 = 150
From the above explanation number of the cubes with 0 faces painted is 150.
From the above explanation number of the cubes with at least 2 faces painted is 44 + 4 = 48.

For least number of cuts 120 = 4 x 5 x 6 i.e number of cuts must be 3, 4 and 5 in three planes in this case number of cubes on a face is either 6 x 5 = 30 or 6 x 4 = 24 or 4 x 5 = 20 cubes . And number of cuboids on an edge is 4 or 5 or 6
Number of cuboids with no face painted is (4 - 2)(5 - 2)(6 - 2) = 2 x 3 x 4 = 24

For least number of cuts 120 = 4 x 5 x 6 i.e number of cuts must be 3, 4 and 5 in three planes in this case number of cubes on a face is either 6 x 5 = 30 or 6 x 4 = 24 or 4 x 5 = 20 cubes . And number of cuboids on an edge is 4 or 5 or 6
To satisfy this case all the cuboids on the edges and corners must have more than one colour on them. And in that case opposite face must have painted in the same colour.
In that case number of cuboids with 3 colours on them = 8
In that case number of cuboids with 2 colours on them = 4 x (2 + 3 + 4 ) = 36
Hence number of cuboids with at least 1 colour on them is 120 - 36 - 8 = 76

For least number of cuts 120 = 4 x 5 x 6 i.e number of cuts must be 3, 4 and 5 in three planes in this case number of cubes on a face is either 6 x 5 = 30 or 6 x 4 = 24 or 4 x 5 = 20 cubes . And number of cuboids on an edge is 4 or 5 or 6
In this case when k is maximum, one particular colour is used on there faces such that any two faces are adjacent to each other. Required number of cuboids will come from edges but not from vertex = 3 + 4 + 5 + 1 = 13

For least number of cuts 120 = 4 x 5 x 6 i.e number of cuts must be 3, 4 and 5 in three planes in this case number of cubes on a face is either 6 x 5 = 30 or 6 x 4 = 24 or 4 x 5 = 20 cubes . And number of cuboids on an edge is 4 or 5 or 6
Maximum number of cuboid with red colour is possible when cube is painted with red colour in 3 sides with minimum number of common edges (which is equal to 2)
Hence required maximum value is 6 (5 + 5 + 4 - 2) = 72
For minimum number of such cuboid Red colour is used only once and minimum number of cubes in that case is 20
Hence required ratio is 72 : 20 = 18 : 5

For least number of cuts 120 = 4 x 5 x 6 i.e number of cuts must be 3, 4 and 5 in three planes in this case number of cubes on a face is either 6 x 5 = 30 or 6 x 4 = 24 or 4 x 5 = 20 cubes . And number of cuboids on an edge is 4 or 5 or 6
In this case we have to use red and green twice and same colour should be on opposite faces then required cube is given by 4 edges (but not corner), maximum number of cubes one edge is 6 - 2 = 4 so required number of cubes is 4 x 4 = 16