jmc

Since the area of the larger circle is $64\pi$ and each circle is divided into two equal areas, the larger shaded area is $\frac{1}{2}$ of $64\pi ,$ or $32\pi .$

Let $r$ be the radius of the larger circle.

Since the area of the larger circle is $64\pi$ and $r>0,$ we have $$\begin{array}{rl}\pi r^2& =64\pi \\ r^2& =64\\ r& =\sqrt{64}=8.\end{array}$$Since the smaller circle passes through the center of the larger circle and just touches the outer circle, by symmetry, its diameter must equal the radius of the larger circle. (In other words, if we join the center of the larger circle to the point where the two circles just touch, this line will be a radius of the larger circle and a diameter of the smaller circle.)

Therefore, the diameter of the smaller circle is $8,$ so its radius is $4.$

Therefore, the area of the smaller circle is $\pi (4^2)=16\pi ,$ so the smaller shaded area is $\frac{1}{2}\times 16\pi$ or $8\pi .$

Therefore, the total of the shaded areas is $32\pi +8\pi =\boxed{40\pi }.$