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.

Part (ii). Uploading (iii) seperately.
'Hence' questions hint at the relatability of the previous part, meaning u cna derive some assistance from the prev part.

I can't remember which theorem is this, but it relates to forming the equation of f(x) = Q(x)D(x) + R(x)
Q -> quotient
D -> Divisor
R -> Remainder

In part i, we found R ==> whatever is left after taking out R has to be divisible by D (i.e. x+1). E.g. 15/4 = 3 R3, after taking out remainder 3 from 15, we get 12, which is completely divisible by 4. If you didnt spot the correlation between the cubic polynomial and x+1, u can also use Long Division or factor theorem to get the same ans.