Volume Enclosed Between a Surface and a Plane

If I understand correctly, your question is to find the volume bounded in the "hump" (call it $R$) where $F > G$. This will involve integrating a function with respect to $x$ and $y$ over some region $R$ corresponding to the points where $F > G$. That is, $F(x, y) > G(x, y)$ if and only if $(x, y) \in R$.

To do this, we just need to find what the ranges of $x$ and $y$ are that define the region $R$. We can observe that boundary of $R$ is defined by the relation $\cos (x) + \cos (y) = 1/2$. Note that both $\cos (x)$ and $\cos (y)$ are both at least $-1/2$ for the equality to hold, since $\cos$ takes on values in the range $[-1, 1]$. Thus, $x, y \in [-2\pi/3, 2\pi/3]$.

Let us calculate the volume of the region where $x, y \geq 0$. Note that this constitutes a quarter of the hump (let's call this portion $Q$). As with $R$, the region $Q$ is defined by the inequality $\cos (x) + \cos (y) \geq 1/2$, so for any fixed $x$, $\cos (y)$ can range from $1$ to $1/2 - \cos(x)$, so $y$ can range from $0$ to $\arccos(1/2 - \cos x)$. The volume of the region $H$ is given by the integral $$\int_H [F(x, y) - G(x, y)] dx dy = \int_0^{2\pi / 3} \int_0^{\arccos(1/2 - \cos x)} [\cos x + \cos y - 1/2] dy dx$$ To get the area of your desired region $R$, you would just have to multiply the result of this integral by four. Unfortunately, this integral does not seem to have a closed-form solution (the $\arccos(1/2 - \cos x)$ causes quite a few problems), so numerical integration seems to be the way to go.

So it is clear that the region of integration reaches its maximum width when either $x$ or $y$ $=0$. Setting $y=0$, $$F(x,0)=\cos(x)+\cos(0)$$ Then, $$F(x,0)=G(x,y) \implies \cos(x)+\cos(0)=\frac{1}{2}$$ Which has solutions at $x=\pm \frac{2\pi}{3}$. Now let's find the bounds for our $y$ values. So, $$F(x,y)=G(x,y) \implies \cos(x)+\cos(y)=\frac{1}{2}$$ So then $$y=\pm \arccos(\frac{1}{2}-\cos(x)).$$ So our integral is $$\int_{-\frac{2\pi}{3}}^{\frac{2\pi}{3}}\int_{-\arccos(\frac{1}{2}-\cos(x))}^{\arccos(\frac{1}{2}-\cos(x))}(\cos(x)+\cos(y)-\frac{1}{2}) \mathrm{d}y\mathrm{d}x \approx 8.26979$$ Computed here