Jordan normal form of $A=\left(\begin{smallmatrix} 0 & 4 & 2 \\ -3 & 8 & 3 \\ 4 & -8 & -2 \\ \end{smallmatrix}\right)$ 
Find a matrix in Jordan normal form that is similar to 
  $$A=\begin{pmatrix}
0 & 4 & 2 \\
-3 & 8 & 3 \\
4 & -8 & -2 \\
\end{pmatrix}$$

The characteristic equation of $A$ is $(\lambda-2)^3=0$, hence, $\lambda=2$ is a eigenvalue of algebraic multiplicity three. So eigenvector corresponds to $\lambda=2$, 
$$(A-2I)=\begin{pmatrix}
-2 & 4 & 2 \\
-3 & 6 & 3 \\
4 & -8 & -4 \\
\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}$$
$$\begin{pmatrix}x\\y\\z\end{pmatrix}=\left\{\begin{pmatrix}1\\0\\1\end{pmatrix},\begin{pmatrix}2\\1\\0\end{pmatrix} \right\}$$For one more eigenvector I tried to get one general eigenvector. $$(A-2I)=\begin{pmatrix}
-2 & 4 & 2 \\
-3 & 6 & 3 \\
4 & -8 & -4 \\
\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}1\\0\\1\end{pmatrix}\text{ or }\begin{pmatrix}2\\1\\0\end{pmatrix}$$ But both of them are inconsistent system and I stuck. Then how to get complete eigenspace to find $P$ such that $A=PJP^{-1}$

The way you are trying to find the third vector is not going to work, as you have demanded the result of $(A-2I)w$ to be one of the two eigenvectors you found. 
I prefer confirming the minimal polynomial, here $(\lambda - 2)^2.$ And $(A - 2 I)^2 = 0.$ So, take any column vector that is not an eigenvalue, for example $w = (0,0,1)^T.$ The eigenvector that is now forced is $v =(A-2I)w,$ so that $v = (2,3,-4).$ CHECK that it really is an eigenvector. Finally, choose an eigenvector $u$ that is not a multiple of $v.$ I like the difference of your two listed eigenvectors, so I chose $u = (1,1,-1)^T$ since the determinant comes out nicely.
We are going to construct $J = Q^{-1} A Q$ having chosen
$$
Q =
\left(
\begin{array}{ccc}
1&2&0 \\
1&3&0 \\
-1&-4 &1 \\
\end{array}
\right)
$$ 
The determinant of $Q$ is one, the inverse is 
$$
Q^{-1} =
\left(
\begin{array}{ccc}
3&-2&0 \\
-1&1&0 \\
-1& 2 &1 \\
\end{array}
\right)
$$
