Determine invariant subspaces imagine that a matrix of an endomorphism has the characteristic polynomial
$(\lambda-2)^2(\lambda-3)$
now i was wondering whether all invariant subspaces can be determined by $0,V$ and $\ker(A-2)^2, \ker(A-2), \ker(A-3)$? or how do I find them? 
 A: No, you miss a few invariant subspaces. One in the non-diagonalizable case. Infinitely many (assuming the field is infinite) in the diagonalizable case.
So $A$ is a $3\times 3$ matrix over a field $K$. By Cayley-Hamilton, $(A-2I)^2(A-3I)=0$ and $K^3=\ker (A-2I)^2\oplus \ker (A-3I)$. Note that $\{0\}\subsetneq \ker (A-2I)\subseteq \ker (A-2I)^2$, with $\dim \ker(A-2I)^2=2$ and $\dim\ker(A-3I)=1$.
Let $T$ be the linear operator whose matrix is $A$ in the canonical basis. Of course it has the same chracteristic polynomial and eigenvalues/spaces as $A$. 
The invariant subspaces of dimension $0$ and $3$ are $\{0\}$ and $K^3$. It remains to determine those of dimension $1$ and $2$.
Let $F$ be such an invariant subspace. In a basis starting with a basis of $F$, the matrix of $T$ becomes block upper triangular. In particular, the upper left block is the matrix of $T_{|F}$ whose characteristic polynomial divides that of $T$.
Case 1: $A$ is diagonalizable, i.e. $\dim \ker (A-2I)=2$ and $\ker (A-2I)=\ker (A-2I)^2$.
If $\dim F=1$, then $F$ is the span of an eigenvector of $T$. So it is either $\ker (A-3I)$, or any one-dimensional susbpace of $\ker (A-2I)$.
If $\dim F=2$, then the characteristic polynomial of $T_{|F}$ is either $(\lambda-2)^2$ or $(\lambda-2)(\lambda-3)$. In the first case, $F=\ker (A-2I)^2=\ker (A-2I)$. In the second case, $F=Kv\oplus \ker(A-3I)$ for any eigenvector $v\in\ker (A-2I)$.
Case 2: $A$ is not diagonalizable, i.e. $\dim\ker(A-2I)=1$.
If $\dim F=1$, now $F=\ker(A-3I)$ or $\ker(A-2I)$.
If $\dim F=2$ and the characteristic polynomial of $T_{|F}$ is $(\lambda-2)(\lambda-3)$, then $F=\ker(A-2I)\oplus\ker(A-3I)$. And if it is $(\lambda-2)^2$, then $F=\dim\ker (A-2I)^2$.
Remark: an easy case, in any dimension, is when $A$ is diagonalizable with pairwise distinct eigenvalues. Then the invariant subspaces are the $2^n$ direct sums built with the eigenspaces.
