Kieran Fitzgerald fro St Stephens Carramar in Australia got us started on the first parts of the problem using an algebraic approach:
a) For the cubic to have no stationary points the derivative cannot equal $0$
($dy/dx \neq 0$).This means that the discriminant must equal less than $0$. For
this to occur, assuming $y=ax^3+bx^2+cx+d$, both $a$ and $c$ must be either + or -
but they cannot be different. $b$ must also be considerably lower than $a$ and
$c$ as $b^2$ must be less than $4ac$. An example would be $y=4x^3-2x^2+3x$.
b) For a cubic to have stationary points at $x=2$ and $5$ the derivative must
equal $(x-2)(x-5)$. This can then be expanded and anti-differentiated to
become $y=x^3/3-3.5x^2+10x+d$ however $d$ can be any number.
c) In this case the derivative must equal $(x+1)(x+d)$. This is because there
will be a stationary point at $x= -1$ and another at $d$. To make $x=-1$ the
local minimum $d$ must be more than $1$ so that the other turning point is a
maximum. An example would be $y=x^3/3+2x^2+3x$.
Thomas from BHASVIC Sixth Form College Brighton used similar reasoning on the last part of the question:
d) We require $\frac{dy}{dx}=3(x+2)(x-4)$ for one max and one min at $x=-2,4$ in some order. Then integrating, $y=x^3-3x^2-24x$. This has the min and max the wrong way round so use transformation $y \rightarrow -y$ to reflect in $x$-axis, giving $$y=-x^3+3x^2+24x$$ as a possible cubic.