a)
I recall you that a list $w_1,w_2,w_3$ of vectors are linearly independent in $V$ if
$$\forall a,b,c \qquad aw_1+bw_2+cw_3 = 0 \quad \implies \quad a=b=c=0.$$
So let $a,b,c$ be such that $au_1+bu_2+cu_3 = 0$, then we have
$$ 0=au_1+bu_2+cu_3=(a+b+c)v_1+(b+c)v_2+cv_3.$$
Now, $v_1,v_2,v_3$ is a basis, so we must have
$$ c=b+c=a+b+c=0.$$
I leave you to show that this implies that $a=b=c = 0$.
b) For this part, it is not clear what you mean exactly by transition matrix. Moreover, note that the coefficients of a matrix depends on the basis in which you express them. I will show you how to find a matrix $P$ such that $Pu_i = v_i$ for $i=1,2,3$ and express this matrix with respect to the basis $B_1$. You should then be able to reuse this example to answer the question.
We have
$$Pu_1 = Pv_1, \quad Pu_2 = Pv_1+Pv_2, \quad Pv_3=Pv_1+Pv_2+Pv_3,$$
and we want $Pu_1 = v_1,Pu_2=v_2$ and $Pu_3=v_3$. This is implies
$$ Pv_1 = v_1, \quad Pv_2 = v_2-Pv_1=v_2-v_1, \quad Pv_3 = v_3-Pv_2-Pv_1=v_3-v_2.$$
Hence, the matrix $P$ has to satisfy,
$$ P\begin{pmatrix}1\\ 0 \\ 0 \end{pmatrix}_{B_1}={\color{red}{\begin{pmatrix}1\\ 0 \\ 0 \end{pmatrix}_{B_1}}},\quad P\begin{pmatrix}0\\ 1 \\ 0 \end{pmatrix}_{B_1}={\color{blue}{\begin{pmatrix}-1\\ 1 \\ 0 \end{pmatrix}}}, \quad P\begin{pmatrix}0\\ 0 \\ 1 \end{pmatrix}_{B_1}={\color{green}{\begin{pmatrix}0\\ -1 \\ 1 \end{pmatrix}_{B_1}}}.$$
It follows that
$$ [P]_{B_1,B_1}=\begin{pmatrix} \color{red}1 & \color{blue}{-1} & \color{green}0 \\ \color{red}0 & \color{blue}1 & \color{green}{-1} \\ \color{red}0 & \color{blue}0 & \color{green}1 \end{pmatrix}.$$
It is now easily checked that
$$ Pu_1 = P\begin{pmatrix}1\\ 0 \\ 0 \end{pmatrix}_{B_1}=\begin{pmatrix}1\\ 0 \\ 0 \end{pmatrix}_{B_1}=v_1, \quad Pu_2 = P\begin{pmatrix}1\\ 1 \\ 0 \end{pmatrix}_{B_1}=\begin{pmatrix}0\\ 1 \\ 0 \end{pmatrix}_{B_1}=v_2, \quad Pu_3 = P\begin{pmatrix}1\\ 1 \\ 1 \end{pmatrix}_{B_1}=\begin{pmatrix}0\\ 0 \\ 1 \end{pmatrix}_{B_1} =v_3,$$
i.e. $Pu_i=v_i$ for $i=1,2,3$.
