transition matrix and coordinate vector Could you please help me solve these quesitons??

Consider the bases S={$u_1, u_2, u_3$}, and T={$v_1, v_2, v_3$}, with 
$u_1$=[-3, 0, -3], $u_2$=[-3, 2, -1], $u_3$=[1, 6, -1], 
$v_1$=[-6, -6, 0],
$v_2$=[-2, -6, 4], $v_3$=[-2, -3, 7] 
(a) Find the transition matrix from $S$ to $T$ 
(b) Using the result in (a), compute the coordinate vector $[w]_T$ where $w$=[-5, 8, -5]
 A: Hint:
If $U$ is the transition matrix from the standard basis to the basis $S$ ( i.e. the matrix that has as columns the vectors $u_i$), and $V$ is  the transition matrix from the standard basis to the basis $T$, than the transition from $S$ to $T$ is given by the matrix $V^{-1}U$. 
A: As noted by @Emilio Novati, the key idea is to connect to the standard basis. 
Vectors will be colored according to the basis membership, and named chromatically:
$$
 \color{blue}{\mathbf{B}\ (standard)}, \qquad
 \color{red}{\mathbf{R}\ (u)}, \qquad
 \color{green}{\mathbf{G}\ (v)}.
$$
$$
\mathbf{R}_{\color{red}{R}\to \color{blue}{A}}=
\color{black}{\left[
\begin{array}{rrr}
 -3 & -3 & 1 \\
 0 & 2 & 6 \\
 -3 & -1 & -1 \\
\end{array}
\right]}
$$
Example: the second vector in the $\color{red}{\mathbf{S}}$ basis has the following coordinates in the $\color{blue}{standard}$ basis
$$
\mathbf{R}_{\color{red}{S}\to \color{blue}{A}}
\color{red}{\left[
\begin{array}{c}
 0 \\
 1 \\
 0
\end{array}
\right]}
=
\color{blue}{\left[
\begin{array}{r}
 -3 \\
  2 \\
 -1
\end{array}
\right]}
$$
You may think of the matrix $\mathbf{R}_{\color{red}{S}\to \color{blue}{A}}$ as an operator which takes a $\color{blue}{blue}$ vector and returns  a $\color{red}{red}$ vector. 
The inverse matrix is a map which connects vectors in the $\color{blue}{standard}$ basis to vectors in the $\color{red}{\mathbf{R}}$ basis:
$$
\mathbf{R}^{-1}_{\color{blue}{B}\to \color{red}{U}}
%
\color{blue}{\left[
\begin{array}{c}
 0 \\
 0 \\
 1
\end{array}
\right]}
=
%
\frac{1}{24}
\left[
\begin{array}{rrr}
 2 & -2 & -10 \\
 -9 & 3 & 9 \\
 3 & 3 & -3 \\
\end{array}
\right]
%
\color{blue}{\left[
\begin{array}{c}
 0 \\
 0 \\
 1
\end{array}
\right]}
=
%
\color{red}{\frac{1}{24}
\left[
\begin{array}{r}
 -10 \\
   9 \\
  -3
\end{array}
\right]}
$$
For the $\color{green}{\mathbf{G}}$ basis, the maps are
$$
\mathbf{G}_{\color{green}{G}\to \color{blue}{A}}=
\color{black}{\left[
\begin{array}{rrr}
 -6 & -6 & 0 \\
 -2 & -6 & 4 \\
 -2 & -3 & 7 \\
\end{array}
\right]}, \qquad
%
\mathbf{G}^{-1}_{\color{blue}{A}\to \color{green}{G}}=
\color{black}{\left[
\begin{array}{rrr}
 -5 &  7 & -4 \\
  1 & -7 & 4 \\
 -1 & -1 & 4 \\
\end{array}
\right]}
$$
Transition from $S$ to $T$
We connect all bases through the hub of the $\color{blue}{standard}$ basis. Start with a vector in the $\color{red}{\mathbf{R}}$ basis, map that to a vector in the $\color{blue}{standard}$ basis, then map that to a vector in the $\color{green}{\mathbf{G}}$ basis:
$$
\color{red}{
\left[
\begin{array}{c}
 x_{1} \\
 y_{1} \\
 z_{1}
\end{array}
\right]}
%
\quad \Longrightarrow \quad
%
\color{blue}{
\left[
\begin{array}{c}
 x_{2} \\
 y_{2} \\
 z_{2}
\end{array}
\right]}
%
\quad \Longrightarrow \quad
%
\color{green}{
\left[
\begin{array}{c}
 x_{3} \\
 y_{3} \\
 z_{3}
\end{array}
\right]}
%
$$
The formal steps are
$$
\begin{align}
%
  \mathbf{R}_{\color{red}{R}\to \color{blue}{B}}
\color{red}{
\left[
\begin{array}{c}
 x_{1} \\
 y_{1} \\
 z_{1}
\end{array}
\right]}
%
&=
%
\color{blue}{
\left[
\begin{array}{c}
 x_{2} \\
 y_{2} \\
 z_{2}
\end{array}
\right]} \\[3pt]
%
\mathbf{G}^{-1}_{\color{blue}{B}\to \color{green}{G}}
\left(
  \mathbf{S}_{\color{red}{R}\to \color{blue}{B}}
\color{red}{
\left[
\begin{array}{c}
 x_{1} \\
 y_{1} \\
 z_{1}
\end{array}
\right]} \right)
%
&=
%
\mathbf{G}^{-1}_{\color{blue}{B}\to \color{green}{G}}
\color{blue}{
\left[
\begin{array}{c}
 x_{2} \\
 y_{2} \\
 z_{2}
\end{array}
\right]}
%
=
%
\color{green}{
\left[
\begin{array}{c}
 x_{3} \\
 y_{3} \\
 z_{3}
\end{array}
\right]}
%
\end{align}
$$
The operator which maps $\color{red}{red}$ vectors to $\color{green}{green}$ is
$$
  \mathbf{X}_{\color{red}{R}\to \color{green}{G}} = 
  \mathbf{G}^{-1}_{\color{blue}{B}\to \color{green}{G}}
  \mathbf{R}_{\color{red}{R}\to \color{blue}{B}} =
%
\frac{1}{24}
\left(
\begin{array}{rr}
 27 & 33 & 41 \\
 -15 & -21 & -45 \\
 -9 & -3 & -11 \\
\end{array}
\right)%
$$
Computation
Turn a $\color{red}{red}$ vector into a $\color{green}{green}$ vector:
$$
\mathbf{X}_{\color{red}{R}\to \color{green}{G}}
\color{red}{
\left[
\begin{array}{r}
 -5 \\
  8 \\
 -5
\end{array}
\right]}
=
\color{green}{
\frac{1}{18}
\left[
\begin{array}{r}
 -57 \\
  99 \\
  57
\end{array}
\right]}
$$
