I assume you want to compute the change-of-basis-matrix from basis B1 to basis B2.
The first step is to write the new basis vectors as a linear combination of the old basis vectors:
$$(1,0,1)=1\cdot(0,0,1)+1\cdot(1,0,0)+0\cdot(0,1,0)$$
$$(1,1,0)=0\cdot(0,0,1)+1\cdot(1,0,0)+1\cdot(0,1,0)$$
$$(0,1,1)=1\cdot(0,0,1)+0\cdot(1,0,0)+1\cdot(0,1,0)$$
The first column in the change-of-basis-matrix is the coordinate of the first new basis vector with respect to the old basis vectors. Repeat this for the second and third column. The you get the following change-of-basis matrix T:
\begin{equation}
\begin{bmatrix}
1 & 0 & 1 \\
1 & 1 & 0 \\
0 & 1 & 1 \\
\end{bmatrix}
\end{equation}
The general formula to transform new coordinates into old coordinates and vice versa:
coordinate of a point with respect to old basis = $T$ multiplied by coordinate in new basis
coordinate of a point with respect to new basis = $T^{−1}$ multiplied by coordinate in old basis
Remark: T translates from the new basis to the old basis!
The matrix $T$, with respect to the basis $v_1= 1 + x + x^2$, $v_2= x + x^2$, and $v_3=x^2$ can be computed directly as follows
$T(v_1)=T(1+x+x^2)=(1-1+1) + (1+1)x + (2-1)x^2=1 + 2x + x^2= v_1+ v_2 - v_3$,
$T(v_2)=T(x+x^2)= 2x + 2x^2 = 2v_2$,
$T(v_3)=T(x^2)= 1 + x = v_1-v_3$
Therefore, the matrix $T=(a_{ij})$ is the matrix which solves the equation
$\begin{pmatrix}
v_1 & v_2 & v_3
\end{pmatrix} T = \begin{pmatrix} v_1 + v_2 - v_3 & 2v_2 & v_1-v_3 \end{pmatrix}$. I'll let you solve explicitly for the values of $T$.
Best Answer
Call $\mathcal E = \{\mathbf e_1, \mathbf e_2, \mathbf e_3\}$ the standard basis for $\Bbb R^3$. Then any vector $\mathbf v \in \Bbb R^3$ can be written as $$\mathbf v = v^1\mathbf e_1 + v^2\mathbf e_2 + v^3\mathbf e_3$$ for some unique triple of numbers $(v^1, v^2, v^3)$ (note: those superscripts are not exponents, they're just indices). The standard basis is particularly useful because it is an orthonormal basis. That means that $\mathbf e_i \cdot \mathbf e_i = 1$ for all $i$ and $\mathbf e_i \cdot \mathbf e_j = 0$ for all $i\ne j$. This property allows a lot of simplifications when you start working through problems.
The matrix representation of the vector $\mathbf v$ with respect to (wrt) $\mathcal E$ is $$[\mathbf v]_{\mathcal E} = \pmatrix{v^1 \\ v^2 \\ v^3}$$
$\mathcal E$ isn't the only basis for $\Bbb R^3$ however. In fact there are an infinite number of bases. Let $\mathcal B= \{\mathcal b_1, \mathcal b_2, \mathcal b_3\}$ be some arbitrary basis of $\Bbb R^3$. Then by definition we can also expand the vector $\mathbf v$ in the basis $\mathcal B$: $$\mathbf v = \nu^1\mathbf b_1 + \nu^2\mathbf b_2 + \nu^3\mathbf b_3$$ where in general $(v^1, v^2, v^3) \ne (\nu^1, \nu^2, \nu^3)$. Likewise the matrix representation of $\mathbf v$ wrt $\mathcal B$ is $$[\mathbf v]_{\mathcal B} = \pmatrix{\nu^1 \\ \nu^2 \\ \nu^3}$$
A linear transformation $T: V \to W$ exists independent of any bases we choose for the spaces $V$ and $W$. However, once we've chosen those bases we can express the action of $T$ on elements of $V$ in matrix form. Say we choose $\mathcal C$ as our basis for $V$ and $\mathcal D$ as our basis for $W$, then the way that $T$ transforms elements $\mathbf v \in V$ is written in matrix form like: $$[T]_{\mathcal D\leftarrow \mathcal C}[\mathbf v]_{\mathcal C} = [T(\mathbf v)]_{\mathcal D}$$
Notice that we have to specify two bases for any linear transformation $T$ before we can express the action of $T$ in matrix form. However, if $T: V \to V$ then most likely we'd like to choose the same basis for both the domain and codomain of $T$ since they are in fact the same space. Thus often when we have a linear tranformation from a space to itself we'll we'll only specify one basis for $T$ and expect that the reader will understand that $[T]_{\mathcal C}$ really means $[T]_{\mathcal C \leftarrow \mathcal C}$.
An identity transformation $I: V\to V$ is one that doesn't change any vector that it acts on. I.e. $$I(\mathbf v) = \mathbf v$$ for all $\mathbf v\in V$. Most of the time when we talk about an identity matrix we'll be talking about the matrix representation of $I$ wrt to the same basis for the domain and codomain as above. It turns out that $$[I]_{\mathcal B} = \pmatrix{1 & 0 & \cdots & 0 & 0 \\ 0 & 1 & \cdots & 0 & 0 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & 1 & 0 \\ 0 & 0 & \cdots & 0 & 1}$$ for any basis $\mathcal B$. However if we decide that we'd like to choose different bases for the domain and codomain of $I$, then the matrix transformation could take the form of any invertible matrix (ask yourself why it must be an invertible matrix). We call the matrix $[I]_{\mathcal C \leftarrow \mathcal B}$ the change of basis matrix from $\mathcal B$ to $\mathcal C$ because it has the property $$[I]_{\mathcal C\leftarrow\mathcal B}[\mathbf v]_{\mathcal B} = [\mathbf v]_{\mathcal C}$$ for all $\mathbf v \in V$.
For instance, using $[\mathbf v]_{\mathcal E}$ and $[\mathbf v]_{\mathcal B}$ from the first section, it is true that $$[I]_{\mathcal B \leftarrow \mathcal E}\pmatrix{v^1 \\ v^2 \\ v^3} = \pmatrix{\nu^1 \\ \nu^2 \\ \nu^3}$$
The change of basis matrix from a basis $\mathcal C = \{\mathbf c_1, \dots, \mathbf c_n\}$ to $\mathcal D = \{\mathbf d_1, \dots, \mathbf d_n\}$ turns out to be $$[I]_{\mathcal D \leftarrow \mathcal C} = \pmatrix{[\mathbf c_1]_{\mathcal D} & \cdots & [\mathbf c_n]_{\mathcal D}}$$ where this is the matrix whose $i$th column is the vector $[\mathbf c_i]_{\mathcal D}$ for all $i$. Confirm this for yourself by figuring out the action of this matrix on a convenient basis of $V$ (hint: try the basis $\mathcal C$).
With that change of basis idea in hand it is clear that we should be able to transform the bases associated with the domain and codomain of a linear transformation $T$ by the formula: $$[T]_{\mathcal D} = [I]_{\mathcal D\leftarrow \mathcal C}[T]_{\mathcal C}[I]_{\mathcal C\leftarrow \mathcal D}$$
Using all of that info above, let's take a look at your question. You're given $[T] = [T]_{\mathcal E}$ and you're given the basis $\mathcal B$ represented in its coordinates wrt to $\mathcal E$. That is, from your question we know that $$[\mathbf v_1]_{\mathcal E} = \pmatrix{1 \\ 0 \\ 1},\quad [\mathbf v_2]_{\mathcal E} = \pmatrix{0 \\ 1 \\ 1},\quad [\mathbf v_3]_{\mathcal E} = \pmatrix{1 \\ 1 \\ 0}$$ This is all we need to find $[T]_{\mathcal B}$. First we construct the change of basis matrix $[I]_{\mathcal E\leftarrow \mathcal B}$. But notice that it's just $$[I]_{\mathcal E\leftarrow \mathcal B} = \pmatrix{[\mathbf v_1]_{\mathcal E} & [\mathbf v_2]_{\mathcal E} & [\mathbf v_3]_{\mathcal E}} = \pmatrix{1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 0}$$ Then of course because $[I]_{\mathcal B\leftarrow \mathcal E}$ undoes the action of $[I]_{\mathcal E\leftarrow \mathcal B}$, we can see that $$[I]_{\mathcal B\leftarrow \mathcal E} = \pmatrix{1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 0}^{-1} = \frac 12\pmatrix{1 & -1 & 1 \\ -1 & 1 & 1 \\ 1 & 1 & -1}$$
Now that you have the change of basis matrices just multiply out $[T]_{\mathcal B} = [I]_{\mathcal B\leftarrow \mathcal E}[T]_{\mathcal E}[I]_{\mathcal E\leftarrow \mathcal B}$ to get $[T]_{\mathcal B}$. It's as easy as that. ;)