CALCULATOR PART
1. The area of R + S is unsigned, meaning you want to find
[tex]\displaystyle\int_a^b\left|f(x)-g(x)\right|\,\mathrm dx[/tex]
where [tex][a,b][/tex] is the interval between the leftmost and rightmost intersections of [tex]f[/tex] and [tex]g[/tex].
First use your calculator to find these intersections:
[tex]\cos x=\dfrac{x+1}3\implies x\approx-3.64,x\approx-1.86,x\approx0.889[/tex]
so that [tex]a=-3.64[/tex] and [tex]b=0.889[/tex]. Now compute the integral using your calculator:
[tex]\displaystyle\int_a^b\left|f(x)-g(x)|\,\mathrm dx\approx1.662[/tex]
2. The volume, using the washer method, is given by the integral
[tex]\displaystyle\pi\int_{-1.86}^{0.889}(|2-g(x)|^2-|2-f(x)|^2)\,\mathrm dx\approx12.078[/tex]
3. A circle of radius [tex]r[/tex] has area [tex]\pi r^2[/tex]; a semicircle with the same radius thus has area [tex]\frac{\pi r^2}2[/tex]. Each cross section of this solid is a semicircle whose diameter is the vertical distance between [tex]f(x)[/tex] and [tex]g(x)[/tex], or [tex]|f(x)-g(x)|[/tex]. In terms of the diameter [tex]d=2r[/tex], the area of each semicircle would be [tex]\frac{\pi d^2}8[/tex]. Then the volume of the solid is
[tex]\displaystyle\frac\pi8\int_{-3.64}^{-1.86}|f(x)-g(x)|^2\,\mathrm dx\approx0.0425[/tex]
NON-CALCULATOR PART
4. The mean value theorem says that for a function [tex]F[/tex] continuous on an interval [tex][a,b][/tex] and differentiable on [tex](a,b)[/tex], there is some [tex]c\in(a,b)[/tex] such that
[tex]F'(c)=\dfrac{F(b)-F(a)}{b-a}[/tex]
If this [tex]F[/tex] happens to be an antiderivative of [tex]f[/tex], then we end up with
[tex]f(c)=\displaystyle\frac1{b-a}\int_a^bf(x)\,\mathrm dx[/tex]
[tex]\cos x[/tex] is continuous and differentiable everywhere, so the MVT applies. We have [tex]F'(x)=f(x)=\cos x[/tex], so the MVT tells us there is some [tex]c\in[0,\pi[/tex] such that
[tex]\cos c=\dfrac{\sin\pi-\sin0}{\pi-0}=0[/tex]
That is, the average value of [tex]f(x)[/tex] on [tex][0,\pi][/tex] is 0. The MVT says there is some [tex]c[/tex] in the interval such that the function takes on the average value itself; this happens for [tex]c=\frac\pi2[/tex].
5. This question seems to be incomplete...
