Calculate the eigenvector (eigenvalue is given) I make this task short and we continue at the step where we know the eigenvalues.
We have the matrix: $\begin{pmatrix}
7-\lambda & 12         & -8\\ 
-2        & -3-\lambda  & 4\\ 
0         & 0          & 3-\lambda 
\end{pmatrix}$
with eigenvalues $\lambda_{1}=1$ and $\lambda_{2}=3$. There isn't a third one because it's $3$, too.

I'd like to know how to calculate the eigenvector just for $\lambda_{1}=1$ correctly.
Insert $\lambda=1$ into the matrix:
$\begin{pmatrix}
6   & 12  & -8\\ 
-2  & -4  & 4\\ 
0   & 0   & 2 
\end{pmatrix}$
We get system of equation:
\begin{align}I:&\quad 6x+12y-8z&=0\\
II:& \quad-2x-4y+4z&=0\\
III:& \quad2z&=0\end{align}
So $z=0$. Insert that in I: $6x+12y=0 \Leftrightarrow x+2y=0 \Leftrightarrow x=-2y$
Now insert $x=-2y$ and $z=0$ in II: $-2(-2y)-4y=0 \Leftrightarrow 4y-4y=0 \Leftrightarrow 0=0$
So we have eigenvector in general $\begin{pmatrix}
x\\ 
y\\ 
z
\end{pmatrix}$. We know that $z=0$ and we know that $x=-2y$. Put that into the matrix: $\begin{pmatrix}
-2y\\ 
y\\ 
0
\end{pmatrix}= y\begin{pmatrix}
-2\\ 
1\\ 
0
\end{pmatrix}$
So the eigenvector for eigenvalue $\lambda_{1}=1$ is $\begin{pmatrix}
-2\\ 
1\\ 
0
\end{pmatrix}$?

Please tell me if I did it correctly (I'm really not sure since $0=0$. It will probably be asked in my exam as well and I have to know it. I have used some online calculators to check if I did it correctly but they sometimes gave strange results.
 A: About the ''$0=0$''
When you ware solving the system you didn't get that $0=0$ and got an redundant result. What you get was that $\forall y \in \mathbb{R}$ it is true that $(-2y,y,0)$ is a possible solution for your system. This is because your eigenvalue problem lead to an system with infinite solutions. So you get a class of solutions to your eigenvalue problem.
I did it correctly ? part of question
Your solution is perfect. You can, for instance, prove it that it is a correct answer just by verifying that, for $\lambda_1 = 1$ and for $\vec{x_1} = (-2,1,0)$ it is true that
$$A\vec{x_1} = \lambda_1\vec{x_1}$$
Where $A$ is 
$$A = \left(\begin{matrix}
7 & 12 & -8 \\
-2 & -3 & 4 \\
0 & 0 & 3
\end{matrix}\right)$$ 
Doing the calculations we get that 
$$A\vec{x_1} = \left(\begin{matrix}
7 & 12 & -8 \\
-2 & -3 & 4 \\
0 & 0 & 3
\end{matrix}\right)\left(\begin{matrix}-2 \\1\\0\end{matrix}\right) = \left(\begin{matrix}-14+12\\+4-3\\0+0+0\end{matrix} \right) = \left(\begin{matrix}-2 \\1\\0\end{matrix}\right) = \vec{x_1} = \lambda_1\vec{x_1}$$
