Part (IV) is mainly a content question. Have to remember that f(x) and it’s inverse are reflected about y=x. Therefore any intersections they have are also found on y=x which is why you can equate either f(x)=x or f’(x)=x to find intersection between f(x) and f’(x).

For part (V), finding value of a is easy by using comparison as shown. The trickier part is finding value of b which requires you to use the fact that for fh to exist, Range (y-values) of h (Rf) must be a proper subset of Domain (x-values) of f (Df).

Given Df is (-∞,3), Rf can be concluded as (-∞,Y). *note the rounded bracket means non inclusive.

It’s very hard to explain in words but if you use the graph drawn on the right, its easier to understand that max value of Y should be 3 (any larger and fh wouldn’t exist as per mentioned earlier) and the x -value should be -2 (as given by domain of h(x)). Given x=-2 and y=3 can solve for value of b.