Your first eigenvector corresponding to [itex]\lambda= 3[/itex] is correct.
However, you have the wrong characteristic equation and "1" is NOT an eigenvalue. You should have seen that when you wrote [itex](A- \lambda I)x= 0[/itex]:
[tex]\begin{bmatrix}2 & 0 \\ 8 & -2\end{bmatrix}\begin{bmatrix}x_1 \\ x_2\end{bmatrix}= \begin{bmatrix} 0 \\ 0 \end{bmatrix}[/tex]
which gives
[tex]\begin{bmatrix}2x_1 \\ 8x_1- 2x_2\end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}[/tex]
which then requires that [itex]2x_1= 0[/itex] and [itex]8x_1- 2x_2= 0[/itex].
From the second [itex]x_2= 4x_1[/itex] but the first says [itex]x_1= 0[/itex] so [itex]x_2= 4(0)= 0[/itex]. There is NO non-trivial vector satisfying this. 1 is NOT an eigenvalue
(I have no idea where you got "[itex]x_1+ x_2= 0[/itex]".)
The characteristic equation is given by
[tex]\left|\begin{array}{cc}3- \lambda & 0 \\ 8 & -1- \lambda\end{array}\right|= 0[/tex]
Which is, of course, simply [itex](3-\lambda)(-1- \lambda)= 0[/itex].