Find the coordinates of $3x^3-2x^2+x+4$ in the base $(3,-1+x,-2+x^2,-3+x^3)$ I did:
$$\alpha_1(3)+\alpha_2(-1+x)+\alpha_3(-2+x^2)+\alpha_4(-3+x^3) = \\
x^3(\alpha_4)+x^2(\alpha_3)+x(\alpha_2)+(3\alpha_1-\alpha_2-2\alpha_3-3\alpha_4)$$
$$\left\{ 
\begin{array}{c}
\alpha_4 = 3 \\
\alpha_3=-2\\
\alpha_2=1 \\
3\alpha_1-\alpha_2-2\alpha_3-3\alpha_4=4
\end{array}
\right. \\
\Leftrightarrow \\
\left\{ 
\begin{array}{c}
\alpha_4 = 3 \\
\alpha_3=-2\\
\alpha_2=1 \\
\alpha_1=2
\end{array}
\right. \\$$
The coordinates are $(2,1,-2,3)$. Is this correct?
Bonus question: Determine the vector $(1,-2,3,-3)$ in the given base.
I did:
$$(1)*(3)+(-2)*(-1+x)+(3)(-2+x^2)+(-3)*(-3+x^3) = \\
3+2-2x-6+3x^2+9-3x^3=\\
8-2x+3x^2-3x^3$$
Is this correct?

Looks like you did the first part right, barring a careless mistake.
For the second part, $x^3-2x^2+3x-3=3a+(-1+x)b+(-2+x^2)c+(-3+x^3)d$.  So solve as above.
$d=1, c=-2, b=3$ and $a=-1/3$.
So $(-1/3, 3, -2, 1)$.
This problem can also be done by inverting the change of basis matrix: $\begin{pmatrix}3&-1&-2&-3\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}$, and multiplying by $(1,-2,3,-3)^t$.
