Parts iii & iv, as promised.
(iii) is a matter of simple differentiation (for stationary pt coordinate) and finding 2nd derivative (for nature of SP). Since f'' = 0, we can conclude that the SP is an inflection point. In case you haven learnt 2nd derivative (Sec 4 syllabus eludes me), you can also find the nature of SP by subbing x-1 & x+1 into dy/dx, where x is the x-coordinate of SP. You wil find that dy/dx is positive for both x-1 and x+1, which point to a inflection point.

(iv) tests your understanding of graphs and derivatives. f(x) is a cubic graph that only has 1 stationary point, and it is NOT a turning point. For any graph to cut the x-axis >1 time, it needs a turning point for the graph to (literally) turn around to intercept the x-axis again.
k is a constant that affects the y-intercept value. Since k doesn't affect the no. of turning pts the graph has, the graph will only cut the x-axis once, regardless of the real value that k takes.