Finding transition matrix between two bases of $V$ 
$V$ is the set of all polynomials with degree $\le1$. Let $S=\{t+1, t-1\}$ and $T=\{t, t-2\}$ be the ordered bases of $V$. What is the transition matrix $P$ from $S$-basis to $T$-basis and the transition matrix $Q$ from $T$-basis to $S$-basis? If $\alpha=5t+1$, what is $[\alpha]_T$ using the transition matrix $P$?

The only answer I got from an expert was this, from there on I need to repeat it for the second part?
Treat $S$ and $T$ as column vectors, and $P$ and $Q$ as $2\times2$ real matrices. So that:
$$S = P T\\
T = Q S$$
By inspection solve for each element, e.g.:
$$S_1 = t+1 = P_{11} T_1 + P_{12}T_2 = P_{11}t + P_{12}(t-2)$$
Now solve for $P_{11}$ and $P_{12}$ the system:
$t = P_{11} t + P_{12} t\\
1 = -2 P_{12}$
so that you get the first line of $P$:
$P_{11} = 1.5\\
P_{12} = -0.5$
It's a $2\times2$ Matrix, so how should I rewrite the answer and proceed further to solve the problem?
 A: I will calculate $P$ as an example and leave the similar calculation of $Q$ to you. $P$ is the transition matrix from $S\to T$, i.e. $P$ takes in a vector in $S$-basis and returns a vector in $T$-basis. Your "expert answer" seems to be telling you the reverse by writing $S=PT$.
The coordinates of the $S$-basis vectors in $T$-basis are found as follows:$$t+1=(3/2)t+(-1/2)(t-2)\\t-1=(1/2)t+(1/2)(t-2)$$So $t+1$ is represented by $[1,0]^t$ in $S$-basis and $[3/2,-1/2]^t$ in $T$-basis and $t-1$ is represented by $[0,1]^t$ in $S$-basis and $[1/2,1/2]^t$ in $T$-basis. Thus$$P\begin{bmatrix}1\\0\end{bmatrix}=\begin{bmatrix}3/2\\-1/2\end{bmatrix}, P\begin{bmatrix}0\\1\end{bmatrix}=\begin{bmatrix}1/2\\1/2\end{bmatrix}$$giving $P=\begin{bmatrix}3/2&1/2\\-1/2&1/2\end{bmatrix}$.
Can you find $Q$ similarly? All you have to do is find the coordinates of the $T$-basis vectors in the $S$-basis and repeat the above procedure.

 You may discover that $Q=P^{-1}$!

For the remaining question, try to express $\alpha=5t+1$ as $k_1(t+1)+k_2(t-1)$, i.e. find the coordinates $[\alpha]_S=[k_1,k_2]^t$. Now you can find $[\alpha]_T=P[\alpha]_S$.
