The Laurent series for $\sin z$ about $z_0 = 0$ is given by it's Taylor series about $z_0 = 0$ because sine is entire. Then, you have $z$ over the Taylor series of $\sin(z)$. Assuming you want to find the Laurent series to find the residue, I will first do everything approximately removing terms that won't contribute to the residue. i.e.
$\frac{z}{\sin^2(z)} = \frac{z}{\left(z-\frac{z^3}{3!}+\ldots\right)\left(z-\frac{z^3}{3!}+\ldots\right)} = \frac{z}{(z^2 - \frac{z^4}{3}+\ldots)} = \frac{z}{z^2\left(1-\frac{z^2}{3}+\ldots\right)} = \frac{1}{z(1-w)}$
where $1-w = 1-\frac{z^2}{3}+\ldots$. Then $w$ is a small term because we are expanding about $z_0=0$. So we can expand $1-w$,
$\frac{1}{1-w} = 1 + w + w^2 + \ldots$,
hence we find
$\frac{z}{\sin^2(z)} = \frac{1}{z}\left(1+w + w^2 + \ldots\right) = \frac{1}{z}(1+ (\frac{z^2}{3} + \ldots))$,
where everything has been valid if we care only about the residue. Hence, $Res\left(\frac{z}{\sin^2(z)} \right) = 1$.
If instead you actually do require a Laurent series, we can let $w = \frac{1}{z}$ and find the Laurent series of
$\frac{(1/w)}{\sin^2(1/w)}$ so we have
$\frac{(1/w)}{\left(\sum_{n=0}^\infty (-1)^n \frac{(1/w)^{2n+1}}{(2n+1)!}\right)^2} = \frac{(1/w)}{\left(\sum_{n=0}^\infty (-1)^n \frac{(1/w)^{2n+1}}{(2n+1)!}\right)\left(\sum_{n=0}^\infty (-1)^n \frac{(1/w)^{2n+1}}{(2n+1)!}\right)}. $
Note that $\sum_{n=0}^\infty (1/w)^{2n+1} = \sum_{m=-\infty}^0 w^{2m+1}$, we use this identity ($m=-n$) to find
$\frac{(1/w)}{\left(\sum_{n=0}^\infty (-1)^n \frac{(1/w)^{2n+1}}{(2n+1)!}\right)^2} = \frac{1/w}{ \left(\sum_{n=-\infty}^0 (-1)^{-n} \frac{w^{2n+1}}{(2(-n)+1)!}\right)\left(\sum_{n=-\infty}^0 (-1)^{-n} \frac{w^{2n+1}}{(2(-n)+1)!}\right)}$.
Next we multiply the series
$\frac{(1/w)}{ \sum_{n=-\infty}^0 \sum_{m=-\infty}^0 (-1)^{n+m}\frac{w^{2(n+m)+2}}{(2(-n)+1)!(2(-m)+1)!}}$
where I have used that $(-1)^n = (-1)^{-n}$ Now let $l=n+m$, yielding
$\frac{(1/w)}{ \sum_{l=-\infty}^0 \sum_{m=-\infty}^l (-1)^{l}\frac{w^{2l+2}}{(2(m-l)+1)!(2(-m)+1)!}} = \frac{1}{ \sum_{l=-\infty}^0 \sum_{m=-\infty}^l (-1)^{l}\frac{w^{2l+1}}{(2(m-l)+1)!(2(-m)+1)!}}$.
And you find the residue when $l=0$, so you find the residue to be 1, again!
First of all, note that$$\frac{z^3+5}{z(z-1)^3}=-\frac5z+\frac6{z-1}-\frac3{(z-1)^2}+\frac6{(z-1)^3}.$$So, what can we do with $-\frac5z$? We have\begin{align}-\frac5z&=-\frac5{1+(z-1)}\\&=-5\left(1-(z-1)+(z-1)^2-(z-1)^3+\cdots\right)\end{align}As you can see, this doesn't matter for compution of the residue, which is $6$, as you wrote.
Best Answer
Write $z = (-\pi/2) + w$. Then $\sin z = \sin (w-\pi/2) = -\cos w$.
Now, you can easily get the beginning of the Taylor expansion of $1 + \sin z$ around $-\pi/2$:
$$1 + \sin z = 1 - \bigl( 1 - \frac{w^2}{2} + \frac{w^4}{4!} - O(w^6)\bigr) = \frac{w^2}{2}\bigl(1 - \frac{w^2}{12} + O(w^4)\bigr)$$
and therefore
$$\begin{align} \frac{z}{1+\sin z} &= \frac{w-\pi/2}{\frac{w^2}{2}\bigl(1 - \frac{w^2}{12} + O(w^4)\bigr)}\\ &= \frac{2w-\pi}{w^2}\bigl(1 + \frac{w^2}{12} + O(w^4)\bigr)\\ &= -\frac{\pi}{w^2} + \frac{2}{w} - \frac{\pi}{12} + \frac{w}{6} + O(w^2). \end{align}$$
Now replace $w$ with $z - (-\pi/2)$.