I love cos 2

Let the expression be $S$ , then we have:

$\begin{aligned} S & = \cos^8 \left( \frac{\pi}{8} \right) + \cos^8 \left( \frac{3\pi}{8} \right) + \color{#3D99F6}{\cos^8 \left( \frac{5\pi}{8} \right)} + \color{#D61F06}{\cos^8 \left( \frac{7\pi}{8} \right)} \\ & = \cos^8 \left( \frac{\pi}{8} \right) + \cos^8 \left( \frac{3\pi}{8} \right) + \color{#3D99F6}{\left[- \cos \left( \frac{3\pi}{8} \right) \right]^8} + \color{#D61F06}{\left[- \cos \left( \frac{\pi}{8} \right) \right]^8} \\ & = 2 \left[ \cos^8 \left( \frac{\pi}{8} \right) + \cos^8 \left( \frac{3\pi}{8} \right) \right] \\ & = 2 \left[ \left( \cos^2 \left( \frac{\pi}{8} \right) \right)^4 + \left( \cos^2 \left( \frac{3\pi}{8} \right) \right)^4 \right] \\ & = 2 \left[ \frac{1}{2^4}\left( 1+ \cos \left(\frac{\pi}{4} \right) \right)^4 + \frac{1}{2^4}\left( 1+ \cos \left(\frac{3\pi}{4} \right) \right)^4 \right] \\ & = 2 \left[ \frac{1}{2^4}\left(1 + \frac{1}{\sqrt{2}} \right)^4 + \frac{1}{2^4}\left(1 - \frac{1}{\sqrt{2}} \right)^4 \right] \\ & = \frac{1}{2^5} \left[\left(\sqrt{2} + 1\right)^4 + \left(\sqrt{2} - 1\right)^4 \right] \\ & = \frac{2}{2^5} \left[(\sqrt{2})^4 + 6(\sqrt{2})^2 + 1 \right] \\ & = \frac{1}{2^4} \left[4 + 12 + 1 \right] \\ & = \frac{17}{16} \end{aligned}$

$\Rightarrow A - B = 17 - 16 = \boxed{1}$

$\color{#3D99F6}{-----------------------------------------}$

On bonus problem

$\begin{aligned} S & = \cos^8 \left( \frac{\pi}{16} \right) + \cos^8 \left( \frac{3\pi}{16} \right) + \color{#3D99F6}{\cos^8 \left( \frac{5\pi}{16} \right)} + \color{#D61F06}{\cos^8 \left( \frac{7\pi}{16} \right)} \\ & = \color{#3D99F6}{\sin^8 \left( \frac{3\pi}{16} \right)} + \cos^8 \left( \frac{\pi}{16} \right) + \color{#D61F06}{\sin^8 \left( \frac{\pi}{16} \right)} + \cos^8 \left( \frac{3\pi}{16} \right) \\ & = \left[ \sin^2 \left( \frac{\pi}{16} \right)\right]^4 + \left[ \cos^2 \left( \frac{\pi}{16} \right)\right]^4 + \left[ \sin^2 \left( \frac{3\pi}{16} \right)\right]^4 + \left[ \cos^2 \left( \frac{3\pi}{16} \right)\right]^4 \\ & = \frac{1}{2^4} \left( \left[1-\cos \left(\frac{\pi}{8} \right)\right]^4 + \left[1+\cos \left(\frac{\pi}{8} \right)\right]^4 + \left[1-\cos \left(\frac{3\pi}{8} \right)\right]^4 + \left[1+\cos \left(\frac{3\pi}{8} \right)\right]^4 \right) \\ & = \frac{1}{2^3} \left( 1 + 6 \cos^2 \left(\frac{\pi}{8} \right) + \cos^4 \left(\frac{\pi}{8} \right) + 1 + 6 \cos^2 \left(\frac{3\pi}{8} \right) + \cos^4 \left( \frac{3\pi} {8} \right) \right) \\ & = \frac{1}{2^3} \left( 2 + 3 \left[1+\cos \left(\frac{\pi}{4} \right) + 1+\cos \left(\frac{3\pi}{4} \right) \right] + \frac{1}{4} \left[ 1+\cos \left(\frac{\pi}{4} \right) \right]^2 + \frac{1}{4} \left[ 1+\cos \left(\frac{3\pi}{4} \right) \right]^2 \right) \\ & = \frac{1}{2^3} \left( 2 + 3 \left[1+ \frac{1}{\sqrt{2}} + 1 - \frac{1}{\sqrt{2}} \right] + \frac{1}{4} \left[ 1+ \frac{1}{\sqrt{2}} \right]^2 + \frac{1}{4} \left[1 - \frac{1}{\sqrt{2}} \right]^2 \right) \\ & = \frac{1}{2^3} \left(2 + 6 + \frac{1}{2} \left[ 1+ \frac{1}{2} \right] \right) \\ & = \frac{1}{8} \left(8 + \frac{3}{4} \right) \\ & = \boxed{\dfrac{35}{32}} \end{aligned}$

Let $x=\sin\frac{\pi}{8}$ and $y=\cos\frac{\pi}{8}$ .

Then $x^2+y^2=1$ and $2xy=\sin\frac{\pi}{4}=\frac{1}{\sqrt{2}}$ , which means $(xy)^2=\frac{1}{8}$ .

Now the question is equivalent to finding $2\left(x^8+y^8\right)$ .

From $x^2+y^2=1$ , we have $x^4+y^4+2(xy)^2=1$ , which means that $x^4+y^4=\frac{3}{4}$ as $(xy)^2=\frac{1}{8}$ . Moving on, $x^8+y^8+2(xy)^4=\frac{9}{16}$ , which means that $x^8+y^8=\frac{17}{32}$ . Hence $2\left(x^8+y^8\right)=\frac{17}{16}$ .