it is possible for a matrix to have fewer leading 1's than rows in reduced row echelon form. in this case, some number of rows at the bottom will be all 0's. the leading 1's may not always be on the diagonal.
for your matrix (a) it is indeed in reduced row echelon form. the 2nd leading 1 in row 2 is indeed rightward of the leading 1 in row 1 (which is always on the diagonal, unless the matrix is the 0-matrix). more importantly, the entire 3rd column is 0's except for the 1 in the 2nd row. the only relevant entry in the last column of matrix (a) is the one in the 3rd row. since the leading 1 of row 2 is in column 3, the first 3 entries of row 3 MUST be 0 for (a) to be in either rref form or echelon form, and the 4th entry must either be 0, or 1 if non-zero. since the 4th entry in row 3 is 0, the entries above it might be any number. if the 4th entry had been 1, in order for the matrix (a) to be in rref form, the entries above it would then have to be 0.
think of it like this: no matter what multiple of row 3 we add/subtract from rows 1 or 2, nothing will change. this matrix is equivalent to the following set of equations:
x = -5w
z = -4w
if you were solving for (x,y,z,w), you could pick any values you like for y and w, then x and z are determined. the matrix has 2 non-zero rows in rref form, and it is no coincidence that you must make 2 choices to solve the system of equations. this number, 2, is the rank of the system (or of the associated matrix).
the main difference between reduced row echelon form and row ecehelon form, is that the numbers above the leading 1 may be non-zero in row echelon form, but must be 0 in reduced row echelon form. in both cases, the entries in a column containing a leading 1 must be 0 below the leading 1 (or else the leading 1 in a subsequent row wouldn't be to the right of some previous leading 1).
in matrix (d) the leading 1 in row 2 is NOT to the right of the leading 1 in row 1. it is 2 places to the LEFT. thus this matrix is neither in reduced row echelon form nor row ecehlon form.