Ah my year’s paper. There’s a lot of logical reasoning/visualising behind a lot of questions in 2015 as I recall. Part (I) is rather simple just draw the graph without modulus and then flip the negative Y area to the positive Y area. Part (II) theres 2 methods to solve can either use algebra and compare or the draw in the line y=3x-1 and then you can see that it cuts the “reflected line” but not the original line since the gradient diverges from y=2x-4 (3>2 so Line with M3 will not meet line with M2 where M is gradient) so can just equate y=3x-1 to y=-2x+4 which is the eqn of the “reflected line”. Part (III) is based a lot on visualisation and logical reasoning; you need to understand how you can get 2 intersections and how there’ll only be 1 intersection. Only if the gradient “converges” with y=2x-4 will you get 2 intersection which means you need a gradient less than 2 so that it meets the graph. However if you have a gradient of 0.5, you will see that the y=0.5x-1 only cuts (2,0) and nothing else which is why M must be bigger than 0.5 but smaller than 2. Remember to write in set notation as that’s what the question asks.