The eigenvalues are correct.
To find the eigenvectors, you need to solve systems of linear equations. Namely, to find eigenvectors of $1$, you need to find all vectors $\mathbf{x}$ for which $(A-(1)I)\mathbf{x}=\mathbf{0}$ (this is equivalent to $(I-A)\mathbf{x}=\mathbf{0}$, but it involves changing fewer entries, so it may be less prone to errors).
$A-I = \left(\begin{array}{cc} 3&\sqrt{2}\\ \sqrt{2}&2\end{array}\right) - \left(\begin{array}{cc}1&0\\0&1\end{array}\right) = \left(\begin{array}{cc}2 & \sqrt{2}\\ \sqrt{2} & 1 \end{array}\right).$
When is $(A-I)\mathbf{x}=\mathbf{0}$? We need to solve the system $\left(\begin{array}{cc} 2 & \sqrt{2}\\ \sqrt{2}&1\end{array}\right)\left(\begin{array}{c}x_1\\x_2\end{array}\right) = \left(\begin{array}{c}0\\0\end{array}\right).$
At this point, you probably know how to solve systems of linear equations. For example, we can use Gaussian elimination on the matrix: $\begin{align*} \left(\begin{array}{cc} 2 & \sqrt{2}\\ \sqrt{2} & 1 \end{array}\right) &\to \left(\begin{array}{cc} 1 & \frac{\sqrt{2}}{2}\\ \sqrt{2} & 1 \end{array}\right) &\text{(divide first row by }2\text{)}\\ &\to \left(\begin{array}{cc} 1 & \frac{\sqrt{2}}{2}\\ 0 & 1 - \sqrt{2}\left(\frac{\sqrt{2}}{2}\right)\end{array}\right) &\text{(subtract }\sqrt{2}\text{ times the first row from row 2)}\\ &=\left(\begin{array}{cc} 1 & \frac{\sqrt{2}}{2}\\ 0 & 0 \end{array}\right). \end{align*}$ The system has infinitely many solutions (as it should, since there have to be eigenvectors associated to the eigenvalue). They are all vectors $(x_1,x_2)^t$ such that $x_1+\frac{\sqrt{2}}{2}x_2 = 0$. That means that we must have $x_1 = -\frac{\sqrt{2}}{2}x_2$. If we set $x_2=\sqrt{2}$, then $x_1=-1$. So one eigenvector is $(-1,\sqrt{2})^t$.
Now you need to do the same thing with $A-4I$, that is, with the matrix $\left(\begin{array}{rr} -1 & \sqrt{2}\\ \sqrt{2} & -2 \end{array}\right).$
(No, $(1,1)$ is not an eigenvectors; note that $A(1,1)^t = (3+\sqrt{2},2+\sqrt{2})^t$, which is not a scalar multiple of $(1,1)$)