**For part 2:**

With 
\begin{equation*} 
y_{\alpha_1}=\begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix},y_{\alpha_2}=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{pmatrix},y_{\alpha_3}=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}
\end{equation*} 
we have \begin{align} 
[y_{\alpha_1},y_{\alpha_2}]&= 0, & (1)\\
[y_{\alpha_1},y_{\alpha_3}]&= -y_{\alpha_2}. & (2)
\end{align}

Then 
\begin{align}
0 &=\delta_{21}\circ \delta_3(v_{(−2,0,2)})=\delta_{21}(a_1y_{\alpha_3}v_{(−2,1,1)},a_2y_{\alpha_1}v_{(−1,−1,−2)})\\
&=(a_1y_{\alpha_3}b_1y_{\alpha_1}^2+a_2y_{\alpha_1}b_2(2y_{\alpha_1}y_{\alpha_3}+y_{\alpha_2}))v_{(0,-1,1)} \\
&=(a_1b_1y_{\alpha_3}y_{\alpha_1}^2+ 2a_2b_2y_{\alpha_1}^2y_{\alpha_3}+a_2b_2y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)}\\
\end{align} 
with $(2)$ follows 
\begin{align}
&=(a_1b_1(y_{\alpha_1}y_{\alpha_3}+y_{\alpha_2})y_{\alpha_1}+2a_2b_2y_{\alpha_1}^2y_{\alpha_3}+a_2b_2y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)} \\
&=(a_1b_1y_{\alpha_1}y_{\alpha_3}y_{\alpha_1}+a_1b_1y_{\alpha_2}y_{\alpha_1}+2a_2b_2y_{\alpha_1}^2y_{\alpha_3}+a_2b_2y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)}
\end{align}
Applying $(2)$ again and additionally $(1)$ we get 
\begin{align*}
&=(a_1b_1y_{\alpha_1}(y_{\alpha_1}y_{\alpha_3}+y_{\alpha_2})+a_1b_1y_{\alpha_1}y_{\alpha_2}+2a_2b_2y_{\alpha_1}^2y_{\alpha_3}+a_2b_2y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)} \\
&=(a_1b_1y_{\alpha_1}^2y_{\alpha_3}+a_1b_1y_{\alpha_1}y_{\alpha_2}+a_1b_1y_{\alpha_1}y_{\alpha_2}+2a_2b_2y_{\alpha_1}^2y_{\alpha_3}+a_2b_2y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)} \\
&=(a_1b_1y_{\alpha_1}^2y_{\alpha_3}+2a_1b_1y_{\alpha_1}y_{\alpha_2}+2a_2b_2y_{\alpha_1}^2y_{\alpha_3}+a_2b_2y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)} \\
&=((a_1b_1+2a_2b_2)y_{\alpha_1}^2y_{\alpha_3}+(2a_1b_1+a_2b_2)y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)} .
\end{align*}

But this would again imply that $a_i, b_i$ are trivial?! So which mistake I did this time...