How do I determine the transformation matrix T of the coordinate transformation from the base E to the base B? In $\mathbb{R}^3$ the canonical basis $E=\left (\mathbf{e_1},\mathbf{e_2},\mathbf{e_3}  \right )$ and $B=\left (\mathbf{b_1},\mathbf{b_2},\mathbf{b_3}  \right )$ with
$\mathbf{b_1}=(1,2,4)^T$, $\mathbf{b_2}=(0,-1,1)^T$ and $\mathbf{b_3}=(2,3,8)^T$.
How do I determine for vector $\mathbf{v}=2\mathbf{e_1}+\mathbf{e_2}+2\mathbf{e_3}$ coordinates $\left \lfloor \mathbf{v} \right \rfloor_E$ and $\left \lfloor \mathbf{v} \right \rfloor_b$ ?
a) $\left \lfloor \mathbf{v} \right \rfloor_E$ =$\begin{pmatrix} 2\\ 1\\ 
2\end{pmatrix}$
$\left \lfloor \mathbf{v} \right \rfloor_b$ =$\begin{pmatrix} -16\\ 6\\  
9\end{pmatrix}$
How do I determine for vector $\mathbf{w}=\mathbf{b_1}+2\mathbf{b_2}+3\mathbf{b_3}$ coordinates $\left \lfloor \mathbf{w} \right \rfloor_E$ and $\left \lfloor \mathbf{w} \right \rfloor_b$ ?
b)
 $\left \lfloor \mathbf{w} \right \rfloor_E$ =$\begin{pmatrix} 7\\ 9\\ 
30\end{pmatrix}$
$\left \lfloor \mathbf{v} \right \rfloor_b$ =$\begin{pmatrix} 1\\ 2\\  
3\end{pmatrix}$
Where I need help?
Determine the transformation matrix T of the coordinate transformation from the base E to the base B, where the old coordinates on E and the new coordinates refer to B. Note: Determine the matrix T such that applies
$\left [ \vec{x} \right ]_E=T\left [ \vec{x} \right ]_B $
I really don't understand how to do it?
d) How can I calculate $\left [ \vec{x} \right ]_b$ from $\left [ \vec{x} \right ]_E$. Take care that $\left [ \vec{x} \right ]_E=\vec{x}$.
 A: Hint:
The matrix
$$ M=
\begin{bmatrix}
1 & 0 & 2\\
2 & -1 & 3\\
4 & 1 & 8
\end{bmatrix}
$$
represents the transformation:
$$
\mathbf{e_1}\to \mathbf{b_1} \qquad \mathbf{e_2}\to \mathbf{b_2} \qquad \mathbf{e_3}\to \mathbf{b_3}
$$
and its inverse:
$$M^{-1}
\begin{bmatrix}
-11 & 2 & 2\\
-4 & 0 & 1\\
6 & -1 & -1
\end{bmatrix}
$$
represents the transformation:
$$
\mathbf{b_1}\to \mathbf{e_1} \qquad \mathbf{b_2}\to \mathbf{e_2} \qquad \mathbf{b_3}\to \mathbf{e_3}
$$
Use $M^{-1}$ to substitute $\mathbf{e_i}$ in the vector $\mathbf{v}$ and  $M$ to substitute $\mathbf{b_i}$ in the vector $\mathbf{w}$

Note that the columns of $M$ are the vectors $\mathbf{b_i}$ in the standard basis, so $M\mathbf{e_i}=\mathbf{b_i}$. In the same way the columns of $M^{-1}$ are the vectors of the standard basis expressed in the basis $\mathbf{b_i}$. So, by linearity, your vector $\mathbf{v}$ that in the standard basis is $\mathbf{v}=2\mathbf{e_1}+\mathbf{e_2}+2\mathbf{e_3}$, in the basis $B$ is:
$$
M^{-1}\mathbf{v}=
\begin{bmatrix}
-11 & 2 & 2\\
-4 & 0 & 1\\
6 & -1 & -1
\end{bmatrix}
\begin{bmatrix}
2\\
1\\
2
\end{bmatrix}=
\begin{bmatrix}
-16\\
-6\\
9
\end{bmatrix}
$$
and the vector $\mathbf{w}$ that in the basis $B$ is $[1,2,3]^T$ , in the standard basis is:
$$
M\mathbf{w}=
\begin{bmatrix}
1& 0 & 2\\
2 & -1 & 3\\
4 & 1 & 8
\end{bmatrix}
\begin{bmatrix}
1\\
2\\
3
\end{bmatrix}=
\begin{bmatrix}
7\\
9\\
30
\end{bmatrix}
$$
A: From what I understand, you are looking for the base transformation matrix from the base $E$ (standard basis) to $B$, i.e. a matrix $T$ such that $Te_1=b_1$, $Te_2=b_2$ and $Te_3=b_3$. This matrix is simply given by $T=(b_1\, b_2\, b_3)$.
Edit
There is nothing to do anymore. The matrix is
\begin{align}
T=\begin{pmatrix}
1 & 0 & 2 \\
2 & -1 & 3 \\
4 & 1 & 8
\end{pmatrix}
\end{align}
You can check that it indeed transforms the basis $E$ to the basis $B$.
A: Let $E = (e_1, e_2, e_3)$ be the old coordinate basis, then any vector $x=(x_1, x_2, x_3)$ can be written as $x = x_1e_1+x_2e_2+x_3e_3$. Let $B = (b_1, b_2, b_3)$ be the new coordinate system. Let $\alpha_1, \alpha_2, \alpha_3$ be the coordinates of $x$ w.r.to the new basis. Then $x$ can be written as $x = \alpha_1b_1+\alpha_2b_2+\alpha_3b_3$. So 
\begin{equation}
\begin{split}
\alpha_1b_1+\alpha_2b_2+\alpha_3b_3 &= x_1e_1+x_2e_2+x_3e_3\\
[b_1, b_2, b_3](\alpha_1, \alpha_2, \alpha_3)^{T} &= [e_1, e_2, e_3] (x_1, x_2, x_3)^{T}\\
(\alpha_1, \alpha_2, \alpha_3)^{T} &= [b_1, b_2, b_3]^{-1} [e_1, e_2, e_3] (x_1, x_2, x_3)^{T}
\end{split}
\end{equation}
Therefore the transformation matrix (from basis $E$ to $B$) is given by $T = [b_1, b_2, b_3]^{-1} [e_1, e_2, e_3]$
