4: Since the cube can be rotated, assume B is fixed to the front face of the cube. Hence, there are 4 possible ways R can be oriented around B. Since there are rotations, it is redundant to multiply by the 6 faces on the cube. Hence 4 ways.
1) That says nothing about the other colors; just that red is either above, below, to the left, or to the right, of the blue. 2) Because those are rotations THEY are redundant. You can count this as $1$ way. But no you have to figure out the different ways to put the remaining four colors.
10: If I map out the cube by drawing 4 horizontal boxes and 1 box below and 1 box above, and assign B to each face, I count the number of ways R can be next to B for each box, and arrive at 10.
24: Similar concept to 4, just multiplied by 6.
Coincidentally that is the correct answer but for entirely the wrong meaning.
Rotation doesn't matter. So you can always place the Blue face toward you. There are four faces adjacent to the Blue and one opposite. If the face opposite is Red that is one way to not do it. All other ways to paint the cube have the red face adjacent.
BUT because rotation doesn't matter those four faces are considered to be the same. So whichever face of those is red we can rotate it to the top.
But now we have rotated the cube so that the red face is on top. That we can not rotate it in any way and keep those two faces the in the same orientation.
There are four faces that need to be painted. Call them A,B,C,D. there are four choices of color for $A$. Once $A$ is chosen there are $3$ left for $B$ and so one. So there are $4*3*2*1 = 24$ to do this.
It might be worth figuring how many ways in total there are to paint the cube.
Either the Red is adjacent to the blue (and there are $24$ ways to do that) or it is opposite the blue. If it opposite then the four remaining sides are all adjecent to both the blue and the red and the cube may be rotated so that any of them may be on top.
One of those four sides must be color 3. Paint one of them color 3 and rotate so color 3 is on top, Red is back, and Blue is facing you. The cube can not be rotated any further. There are $3$ faces left. Call them $A,B,C$. There are $3$ choices of color for face A, and after that $2$ for face $B$ and so on. So there are $3*2*1 = 6$ ways to have the Red face Opposite the Blue face.
So there are $24+6 = 30$ ways to pain the cube of which $\frac 45$ of them have the Red adjacent to Blue.