Note that, if G is an abelian group, then there is no need to speak about "left" and "right" cosets (for (a), (b) and (d) specifically).
(a) Z / 3Z = { 0 ; 1 ; 2 }
where :
0 = { integers that are multiples of 3 }
1 = { integers of the form 3k + 1 }
2 = { integers of the form 3k + 2 }
(b) Let M in GL4(R).
The left coset with respect to M is : M.SL4(R) = { M.N, with N in SL4(R) }
Let's show that : M.SL4(R) = { P in GL4(R), det(P) = det(M) }.
If P is in M.SL4(R), then P = M.N for some N in SL4(R). Hence, det(P) = det(M.N) = det(M).det(N) = det(M).
Now if P is such that det(P) = det(M), then P = M.(M⁻¹.P), and det(M⁻¹.P) = det(M⁻¹).det(P) = det(P)/det(M) = 1, so M⁻¹.P is in SL4(R).
Similarly, you can show that the right coset SL4(R).M is the same subset of GL4(R).
(use the decomposition : P = (P.M⁻¹).M )
Note that, since the left and right cosets are the same, you can define the quotient group, which is in bijection with R* (the set of possible different determinants).
(c) R* / {-1;1} = { { a ; -a }, a > 0 }
Two elements a and b of R* are in the same coset if and only if : a/b ∈ { -1 ; 1 }, ie : if and only if b = - a.
So you can define a coset by chosing the positive element, ie : R* / { -1 ; 1 } ~ R+* ( or ~ R-* if you prefer to choose the negative one).
(d) Q / Z = { q ∈ Q / 0 ≤ q < 1 }
Let [.] be the floor function.
For every q ∈ Q, 0 ≤ q - [q] < 1, hence, with q' = q - [q], you can always choose, in any coset, an element of [0;1[.
Reciprocally, if q1 and q2 are in [0;1[, they cannot lie in the same coset, since 0 < q2 - q1 < 1 : q2 - q1 cannot be an integer.