How to find Laurent series Expansion

The problem is that if you use $\frac 1{1-z} = \sum z^n$ you are essentially writing the Laurent expansion in a neighborhood of $0$. Since you want powers of $z-1$, it means that you want an expansion in a neighborhood of $1$!

Hence the $\frac {1}{z-1} = (z-1)^{-1}$ term is already "good" (just like $\frac 1z$ would be in a Laurent expansion in a neighborhood of $0$).

You have to expand in a neighborhood of $1$ the expression $\frac z{z-3}$. You can set $t = z - 1 \implies z = t + 1$ so your expression becomes $\frac{t+1}{t-2}$.

Now use partial fractions, and find the Laurent series in a neighborhood of $0$ with respect to $t$ of $\frac{t+1}{t-2}$ (This means you can use the geometric series formula), substitute back $t = z-1$ and divide everything by $z-1$ to get the final result.

No need for contour integrals, just give a name to the quantity you want a Laurent series in, and expand. So with $x=z-1$: $$ \frac z{(z-1)(z-3)}=\frac{x+1}{x(x-2)} =x^{-1}\left(1-\frac3{2-x}\right) \\=x^{-1}\left(1-\frac32\sum_{i\geq0}\bigl(\frac x2\bigr)^i\right) =-\frac12x^{-1}+\sum_{i\geq0}\frac{-3}{4\times2^i}x^i. $$ You can now substitute $x:=z-1$ if you like.

First, yes. The same function can have different Laurent series, depending on the center of annulus in question.

Consider now the given function. Clearly, the question will be solved with ease once we find the series of $$g(z)=\frac{z}{z-3}.$$Note that$$g(z)=1+\frac{3}{z-3}=1+\frac{3}{(z-1)-2}=1+\frac{3/2}{\frac{z-1}{2}-1},$$and the last expression can be represented as the sum of a geometric series.