Let $f(x)=(2x-3)^3$; then after multiplying out we have $f(x)=-27+54x-36x^2+8x^3$, so if we write $f$ as a power series, $f(x)=\sum_{k\ge 0}a_kx^k\;,$ we must have $a_0=-27,a_1=54,a_2=-36,a_3=8$, and $a_k=0$ for $k\ge 4$.
For the second problem you have $f(x)=\frac{x^3}{1-x^2}\;.$ Deal with the denominator first. You should know the simple generating function that gives the sum of a geometric series:
$\frac1{1-x}=\sum_{k\ge 0}x^k\;.$ Replace $x$ by $x^2$, and you have
$\frac1{1-x^2}=\sum_{k\ge 0}\left(x^2\right)^k=\sum_{k\ge 0}x^{2k}\;.$
Finally, multiply by $x^3$:
$\frac{x^3}{1-x^2}=x^3\sum_{k\ge 0}x^{2k}=\sum_{k\ge 0}x^{2k+3}\;.$
Think of this now as $\sum_{k\ge 0}a_kx^k$; what are the coefficients $a_k$? It may help to write out a few terms of this sum:
$\frac{x^3}{1-x^2}=x^3+x^5+x^7+x^9+\ldots\;.$
Clearly $a_k=0$ when $k$ is even and when $k=1$, and $a_k=1$ when $k$ is an odd number greater than $1$. Thus, the associated sequence is $\langle 0,0,0,1,0,1,0,1,0,\dots\rangle\;.$