Let's solve the problem step by step.

We are given the vector $v = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$ and the matrix $A = 2 \begin{bmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{bmatrix}$. We need to find $A^8v$.

Step 1: Simplify Matrix A

Start by calculating the simplified form of matrix $A$: $A = 2 \begin{bmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{bmatrix} = \begin{bmatrix} \sqrt{2} & -\sqrt{2} \\ \sqrt{2} & \sqrt{2} \end{bmatrix}.$

Step 2: Analyze the Matrix $A$

To make computing higher powers of the matrix easier, we should check if the matrix can be diagonalized or if it has a special form like a rotation or reflection matrix. Let's calculate the eigenvalues and eigenvectors.

Calculate the characteristic equation: $\det(A - \lambda I) = \det \begin{bmatrix} \sqrt{2} - \lambda & -\sqrt{2} \\ \sqrt{2} & \sqrt{2} - \lambda \end{bmatrix} = (\sqrt{2} - \lambda)^2 - (-\sqrt{2})(\sqrt{2}) = \lambda^2 - 2\sqrt{2}\lambda.$ Simplifying this will provide insights about the matrix's behavior.

Since the matrix relates to rotation, we can infer that its power to even numbers will yield cyclic results.