Find the change of coordinate matrices:
Wherein B is the standard basis for P2 $$B' = (t^2+2,t+3,t^2+t+1) \\B" = (2t^2+t+1, t^2, 2t+1) \\ B= (t^2,t,1) $$ $$P_{B'B}$$ means the transformation for the standard basis to B'
$$ B' = \{\,t^2 + 2\,,\; t+3\,,\; t^2+t+1\,\}$$
$$P_{BB'} \\ P_{B'B} \\P_{BB"}\\ P_{B"B} \\ P_{B'B"} \\ P_{B"B'}$$
The answers I got from respectively(first four) were: They are all wrong. What are the steps to do this kinds of questions?
$$\begin{pmatrix} 1 &0&2 \\ 0 & 1 & 3 \\2 &3&1 \end{pmatrix}$$
$$\begin{pmatrix} 1 &0&0 \\ 0 & 1 & 0 \\2 &0&1 \end{pmatrix}$$
$$\begin{pmatrix} 2 &0&0 \\ 1 & 1 & 0 \\1 &0&1 \end{pmatrix}$$
$$\begin{pmatrix} 2 &1&1 \\ 1 & 0 & 0 \\0 &0&1 \end{pmatrix}$$
Let's look at the vectors of $\mathcal{B}'$, represented as coordinate vectors in two ways. If $\mathcal{B}$ is the basis that we use for coordinates, then you have $$\begin{bmatrix}1\\0\\2\end{bmatrix},\begin{bmatrix}0\\1\\3\end{bmatrix},\begin{bmatrix}1\\1\\1\end{bmatrix}$$ but if $\mathcal{B}'$ is the basis that we use for coordinates, then you just have $$\begin{bmatrix}1\\0\\0\end{bmatrix},\begin{bmatrix}0\\1\\0\end{bmatrix},\begin{bmatrix}0\\0\\1\end{bmatrix}$$ When you ask for $P_{\mathcal{B}\mathcal{B}'}$, you are therefore looking for a matrix that satisfies $$P_{\mathcal{B}\mathcal{B}'}\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=\begin{bmatrix}1&0&1\\0&1&1\\2&3&1\end{bmatrix}$$ so there you go; $P_{\mathcal{B}\mathcal{B}'}$ is apparent.
The next one on your list is the inverse of $P_{\mathcal{B}\mathcal{B}'}$.
The two after that can be found similarly.
For the last two, consider that $P_{\mathcal{B}''\mathcal{B}}P_{\mathcal{B}\mathcal{B}'}=P_{\mathcal{B}''\mathcal{B}'}$.