been a while, I'd wait for someone to confirm my answers or tell me I'm full of it. Anyway...

a)Inverse of f switch x and y values. so you get a curve in the same quadrant with it curving up toward x = infinity. So evaluating x=1 is like finding x where y=1 on the original graph. Which is 0.

b)

(f^-1 o g)(1)=> Well g(1) = 2, and f^-1(2) =1(remember its like finding x where y=2 in f).

(g o f )(1)=> f(1)=2, g(2) = 2^2 + 1 =5

c)The graph will start at (-3,0), curve down through points (-2,-1) & (-1,-3), ad end at point (0,-7).

To explain...

1st)The f(x+3) is like moving the x-axis 3 to the right. Which has the effect of moving your graph 3 to the left. Making the domain [-3,0] instead of [0,3].

2nd)The f(x+3) is subtracted from something so basically whatever y value comes out of it is made negative. This has the effect of flipping the graph about the x-axis. So at this point it curves down from (-3,-1) to (0,-8). So the range at this point is [-1,-8]

3rd)The previous step was subtracted from 1, so we add 1 to all the y values which moves the graph up 1. So the final range is [-0,-7]

d)see explanations to part c