If $\mathbf{x}$ is an eigenvector of $A$ with eigenvalue $\lambda$, then $A\mathbf{x}=\lambda\mathbf{x}$ and $(A-\lambda I)\mathbf{x}=\mathbf{0}$.
First, find the eigenvector corresponding to the eigenvalue $λ=\frac{7+\sqrt{17}}{2}$:
$$\begin{align*} &\quad\quad\quad\quad\left(\begin{array}{c|c}
A-\lambda I & 0
\end{array}\right)\quad\quad\text{insert your }A\text{ and }\lambda\\ &=\left(\begin{array}{cc|c}
4-\tfrac{7+\sqrt{17}}{2} & 2 & 0 \\
2 & 3-\tfrac{7+\sqrt{17}}{2} & 0
\end{array}\right)\quad\quad\text{compute the differences}\\ &\implies \left(\begin{array}{cc|c}
\tfrac{1-\sqrt{17}}{2} & 2 & 0 \\
2 & \tfrac{-1-\sqrt{17}}{2} & 0
\end{array}\right)\quad\quad\text{multiply the first row by }\tfrac{4}{1-\sqrt{17}}\\ &\implies \left(\begin{array}{cc|c}
2 & \tfrac{8}{1-\sqrt{17}} & 0 \\
2 & \tfrac{-1-\sqrt{17}}{2} & 0
\end{array}\right)\quad\quad\text{multiply the first fraction by }1+\sqrt{17}\\ &\implies \left(\begin{array}{cc|c}
2 & \tfrac{8(1+\sqrt{17})}{-16} & 0 \\
2 & \tfrac{-1-\sqrt{17}}{2} & 0
\end{array}\right)\quad\quad\text{simplify the first fraction}\\ &\implies \left(\begin{array}{cc|c}
2 & \tfrac{-1-\sqrt{17}}{2} & 0 \\
2 & \tfrac{-1-\sqrt{17}}{2} & 0
\end{array}\right)\quad\quad\text{subtract the first row from the second}\\ &\implies \left(\begin{array}{cc|c}
4 & -1-\sqrt{17} & 0 \\
0 & 0 & 0
\end{array}\right)\quad\quad\text{deduce the solution}\\ &\implies \mathbf{x}=k\pmatrix{1+\sqrt{17}\\4}\end{align*}$$
Now, normalize it by
$$\hat{\mathbf{x}}=\frac{\mathbf{x}}{||\mathbf{x}||}$$
and do the same thing for the second eigenvalue.