For part (iii), the mistake is in the 3rd integral where you integrated x^(3/2) from 0 to 1. You correctly put a negative sign in front of the integral because the area is below the x-axis. The fact that you have added a negative sign means you should get positive area, but you notice that you obtained area of -2/5 for this portion.

When the curve is symmetrical across the x-axis like this, you need to split the equation to 2 parts. y=+x^(3/2) gives you the part of the curve above the x-axis, while y=-x^(3/2) gives you the part of the curve below the x-axis. The curve you should integrate here is y=-x^(3/2), from x=0 to x=1, and this gives area of +2/5 instead of -2/5. Combining this with the other terms, you get the correct total area of 2.7.

Alternatively, instead of integrating with respect to x, it may be easier if you integrate with respect to y. Treating it as a case of area between 2 curves, you just integrate {(y+4)/3 - y^(2/3)] from y=-1 to y=8.