b) let g \in G and write g as its (unique?) product under the generators a,b. Associate s with a and t with b. Then for each element in G there is a unique function R^2 to R^2 which is defined as the composition of the functions s and t replacing a and b in the factorisation of g. i.e. ig g = aba^2b^(-1) then the function is sts^2t^(-1). Then the action on R^2 is defined for g\in G by applying the associated function to (x,y).
This action is free and properly dis-continious (take B((x,y),1/4) as your open neighbourhood), so then the map R^2 to R^2/G is a covering map, and the orbit space R^2/G is homeomorphic to the klein bottle (I can only prove this by drawing a picture if you sketch out what the quotient space looks like in R^2 though I think it is clear).
c) s^2(x,y) = (x, y+1). So then it should be clear that by defining the group action of H on R^2 as in the previous part is free and properly discontinuous . Then the map R^2 to R^2/H is a covering space, and further H is isomorphic to the fundamental group of R^2/H as R^2 is simply connected (result in lectures). But the identification space identifies (x,y) with (x+n,y+m) for n,m integers and so it is clear (draw a picture) that this is homeomorphic to the torus. So R^2/H = T, but the fundamental group of T is ZxZ the product of the infinite cyclic group.
d) I think the map you want maps you want does this:
b ^ ^ b
Covering map :
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b ^ \/b ^ b
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I hope that picture clears it up?
Edit: Piece of crap doesn't let me draw
Ok well if you have the torus as a (unit) square with opposite edges identified without a twist, the take the following identification for (x,y) with x<= 1/2 sticks to (x +1/2, 1-y). This gives you a two-sheeted cover.
Half of the top 10 aren't even RG...