Let $$f(z)=\frac{\sinh(z)}{(z-1)^4}.$$
First I do the following:
$$\sinh(z)=\sinh((z-1)+1)=\sinh(z-1)\cosh(1)+\cosh(z-1)\sinh(1)$$
The expansions of $\sinh(z)$ and $\cosh(z)$ are $$\begin{align*} \sinh(z) &= z+\frac{z^3}{3!}+\frac{z^5}{5!}+\cdots \\ \cosh(z) &= 1+\frac{z^2}{2!}+\frac{z^4}{4!}+\cdots \end{align*}$$
Then $$\begin{align*} \sinh(z-1) &= (z-1)+\frac{(z-1)^3}{3!}+\frac{(z-1)^5}{5!}+\cdots \\ \cosh(z-1) &= 1+\frac{(z-1)^2}{2!}+\frac{(z-1)^4}{4!}+\cdots \end{align*}$$
Now let's see what $$\begin{align*} \frac{\cosh(1)\sinh(z-1)}{(z-1)^4} &= \cosh(1) \biggl( \frac{1}{(z-1)^3}+\frac{1}{3!(z-1)}+\frac{(z-1)}{5!}+\cdots \biggr) \\ \frac{\sinh(1)\cosh(z-1)}{(z-1)^4} &= \sinh(1) \biggl( \frac{1}{(z-1)^4}+\frac{1}{2!(z-1)^2}+\frac{1}{4!}+\cdots \biggr) \end{align*}$$
Rewriting these last two equalities we have $$\begin{align*} \frac{\cosh(1)\sinh(z-1)}{(z-1)^4} &= \sum_{n=1}^{\infty} \frac{\cosh(1)}{(2n-1)!}(z-1)^{2n-5} \\ \frac{\sinh(1)\cosh(z-1)}{(z-1)^4} &= \sum_{n=0}^{\infty} \frac{\sinh(1)}{(2n)!}(z-1)^{2n-4} \end{align*}$$
Hence the Taylor and Laurent series of $f(z)$ is $$ f(z) = \sum_{n=1}^{\infty} \frac{\cosh(1)}{(2n-1)!}(z-1)^{2n-5} + \sum_{n=0}^{\infty} \frac{\sinh(1)}{(2n)!}(z-1)^{2n-4} $$
Where we have $$ a_{n} = \frac{\cosh(1)}{(2n-1)!} \quad \text{and} \quad b_{n} = \frac{\sinh(1)}{(2n)!}. $$
But I don't know if it's okay. I feel a bit confused regarding these series. Could you tell me if I'm okay? or in another case, could you give me any suggestions?
Your work is good but you can make life a bit easier starting with $z=t+1$ (this is what you implicitly did) $$\frac{\sinh (z)}{(z-1)^4}=\frac{\sinh (t+1)}{t^4}$$ Expanding as you did $$\sinh (t+1)=\sinh (1) \cosh (t)+\cosh (1) \sinh (t)$$ Now, using the expansions of $\cosh (t)$ and $\sinh (t)$ around $t=0$ you then have $$\frac{\sinh (t+1)}{t^4}=\sum_{n=0}^\infty a_n\, t^{n-4}$$ where $$a_{2n+1}=\frac {\cosh(1)}{(2n+1)!}\qquad \text{and} \qquad a_{2n}=\frac {\sinh(1)}{(2n)!}$$ which is your result (just make $t=z-1$).