This is how I've done so far: $$ f(z)=(z^2-1)\mathrm{cos}\frac{1}{z+i} \\w=z+i \;\;\;\; \Rightarrow \;\;\;\; z=w-i \\f(w)=((w-1)^2-1)\mathrm{cos}\frac{1}{w}=(w^2-2wi-2)\mathrm{cos}\frac{1}{w} \\f(w)=w^2\cdot\mathrm{cos}\frac{1}{w}-2wi\cdot\mathrm{cos}\frac{1}{w}-2\cdot\mathrm{cos}\frac{1}{w}= \\= w^2\cdot\sum_{n=0}^{\infty}\frac{(-1)^n}{w^{2n}(2n)!}-2wi\cdot\sum_{n=0}^{\infty}\frac{(-1)^n}{w^{2n}(2n)!}-2\cdot\sum_{n=0}^{\infty}\frac{(-1)^n}{w^{2n}}= \\=\sum_{n=0}^{\infty}\frac{(-1)^n}{w^{2n-2}(2n)!}-2i\cdot\sum_{n=0}^{\infty}\frac{(-1)^n}{w^{2n-1}(2n)!}-2\sum_{n=0}^{\infty}\frac{(-1)^n}{w^{2n}(2n)!}= \\=w^2-\sum_{n=1}^{\infty}\frac{(-1)^n}{w^{2n-2}(2n)!}-2\cdot\sum_{n=0}^{\infty}\frac{(-1)^n}{w^{2n}(2n)!}-2i\cdot\sum_{n=0}^{\infty}\frac{(-1)^n}{w^{2n-1}(2n)!}= \\=w^2-\sum_{n=0}^{\infty}\frac{(-1)^{n+1}}{w^{2n}(2n+2)!}-2\cdot\sum_{n=0}^{\infty}\frac{(-1)^n}{w^{2n}(2n)!}-2i\cdot\sum_{n=0}^{\infty}\frac{(-1)^n}{w^{2n-1}(2n)!} \\ f(w)=w^2-\sum_{n=0}^{\infty}\left (\frac{(-1)^{n+1}}{(2n+2)!}-2\frac{(-1)^n}{(2n)!}\right )\frac{1}{w^{2n}}-2i\cdot\sum_{n=0}^{\infty}\frac{(-1)^n}{w^{2n-1}(2n)!} $$
So finally Laurent series is: $$ f(z)=(z+i)^2-\sum_{n=0}^{\infty}\left (\frac{(-1)^{n+1}}{(2n+2)!}-2\frac{(-1)^n}{(2n)!}\right )\frac{1}{(z+i)^{2n}}-2i\cdot\sum_{n=0}^{\infty}\frac{(-1)^n}{(z+i)^{2n-1}(2n)!} $$
Still don't know how to find residue $Res[f(z), -i]$.
Thank you very much for help in advance!
So if anyone ever need this, here is how this is done:
We found Laurent series for given function, which is in this case the last expression: $$ f(z)=(z+i)^2-\sum_{n=0}^{\infty}\left (\frac{(-1)^{n+1}}{(2n+2)!}-2\frac{(-1)^n}{(2n)!}\right )\frac{1}{(z+i)^{2n}}-2i\cdot\sum_{n=0}^{\infty}\frac{(-1)^n}{(z+i)^{2n-1}(2n)!} $$
Now the residue $Res[f(z), z_{0}]$ is coefficient $b_1$ beside $(z-z_{0})^{-1}$ in Laurent series expansion, so in our case we need the coefficient standing beside $(z+i)^{-1}=\frac{1}{(z+i)^{1}}$. However we have three expressions in our Laurent expansion. The first one is $(z+i)^2$ which never can be $(z+i)^{-1}$ obviously. Similar is for the second expression, since it summarize terms by even numbers ($2n$). So the residue has to be found in the last expression. To get $\frac{b_1}{(z+i)^{1}}$, clearly has to be $2n-1=1$, so we will find the residue for $n=1$ (which is $\frac{(-1)^1}{(z+i)^12!}$).
Finally: $$Res[f(z), -i]=-2i\cdot\frac{(-1)^1}{2!}=i$$