I tried to calculate the Laurent series of these functions but I have no way to verify my answers.
i) $$ \begin{align} f(z)=\frac{e^{z^2 }-1}{z^4}, \mathbb{D}=\mathbb{C} \backslash \{0\} \end{align} $$
We know that $ e^z=\sum_{n=0}^\infty(\frac{1}{n!})z^n $ , hence $$ f(z)=\frac{e^{z^2 }-1}{z^4}=\frac{1}{z^4}(-1+e^{z^2})=\frac{1}{z^4}(-1+\sum_{n=0}^\infty(\frac{1}{n!})(z^2)^n)=\\ =\frac{1}{z^4}(-1+\sum_{n=0}^\infty(\frac{1}{n!})z^{2n})=\frac{1}{z^4}\sum_{n=1}^\infty(\frac{1}{n!})z^{2n}=\\ \sum_{n=1}^\infty(\frac{1}{n!})z^{2n-4} $$
ii)
$$ \begin{align} f(z)=\frac{Logz}{(z-1)^2}, \mathbb{D}=\{ z\in \mathbb{C}:0<|z-1|<1\} \end{align} $$
We know that $Logz=-\sum_{n=1}^\infty\frac{(-1)^n}{n}(z-1)^n$, for $|z-1|<1$ ,hence $$ f(z)=\frac{Logz}{(z-1)^2}=\frac{1}{(z-1)^2}\left(- \sum_{n=1}^\infty\frac{(-1)^n}{n}(z-1)^n \right)= \sum_{n=1}^\infty\frac{(-1)^n}{n}(z-1)^{n-2} $$
iii) $$ \begin{align} f(z)=z^6\cos^2{z^{-2}},z_0=0, \mathbb{D}=\mathbb{C} \backslash \{0\} \end{align} $$
We know that $$\cos z=\sum_{n=0}^\infty\frac{(-1)^n}{(2n)!}z^{2n}$$ therefore $$\cos(\frac{1}{z})=\sum_{n=0}^\infty\frac{(-1)^n}{(2n)!}(\frac{1}{z})^{2n}$$ and $$\cos(\frac{1}{z^2})=\sum_{n=0}^\infty\frac{(-1)^n}{(2n)!}(\frac{1}{z^2})^{2n}$$ We also know that when we a function to the power of 2, its correspondent series will be the derivative of the original series, so: $$\cos^2(\frac{1}{z^2})=1+\sum_{n=1}^\infty\frac{(-1)^n}{(2n)!}2^{2n-1}(\frac{1}{z^2})^{2n}$$ (actually my result here was slightly diferrent, but wolframalfa returned this)
Finally, $$f(z)=z^6+ \left( 1+\sum_{n=1}^\infty\frac{(-1)^n}{(2n)!}2^{2n-1}(\frac{1}{z^2})^{2n} \right)$$
iv) $$f(z)=\frac{1-2z}{z^2-z},\mathbb{D}=\{ z\in \mathbb{C}:|z-1|>1\} $$
$$f(z)=\frac{1-2z}{z^2-z}=\frac{1-2z}{z(z-1)}=-\frac{1}{z}-\frac{1}{z-1}=-\frac{1}{z}+\frac{1}{1-z}==-\frac{1}{z}+\sum_{n=0}^\infty z^n$$
v) $$f(z)=\frac{1}{1+z^2} , \mathbb{D}=\{ z\in \mathbb{C}:1<|z-2i|<3\} $$
$$f(z)=\frac{1}{1+z^2}=\frac{1}{(i-z)(i+z)}=\frac{1}{(2i)}\frac{1}{(i-z)}+\frac{1}{(2i)}\frac{1}{(i+z)}$$
Let $z=2i+u$ so $$\frac{1}{(2i)}\frac{1}{(i-(2i+u))}+\frac{1}{(2i)}\frac{1}{(i+2i+u)}=\frac{1}{(2i)}\frac{1}{(-i-u)}+\frac{1}{(2i)}\frac{1}{(3i+z)}=\\ \frac{1}{(2i)}\frac{1}{u}\frac{-1}{(1-\frac{1}{iu})}+\frac{1}{(2i)}\frac{1}{3i}\frac{1}{(1-\frac{iu}{3})}=\\ \frac{1}{(2i)}\frac{1}{u}\sum_{n=0}^\infty(\frac{1}{iu})^n-\frac{1}{(6)}\sum_{n=0}^\infty(\frac{iu}{3})^n=\\ \frac{1}{(2i)}\frac{1}{(z-2i)}\sum_{n=0}^\infty(\frac{1}{i(z-2i)})^n-\frac{1}{(6)}\sum_{n=0}^\infty(\frac{i(z-2i)}{3})^n $$
If anyone will take the time to verify my results and point my errors I will be grateful because I have no other means of checking their validity. Sorry for the long post.
You first computation is correct. The second is mostly correct, only at the end you dropped a minus sign. (It might have been better to absorb it into the sign in the sum and write $(-1)^{n-1}$ there rather than $(-1)^n$.)
In the third,
is wrong, consider for example polynomials to see that that generally isn't the case.
The square of $\sum\limits_{n=0}^\infty a_n w^n$ is the Cauchy product of the series with itself,
$$\sum_{n=0}^\infty \left(\sum_{k=0}^n a_k a_{n-k}\right) w^n,$$
and $\sum\limits_{k=0}^n a_k a_{n-k}$ rarely is equal to $(n+1)a_{n+1}$. For the cosine, a trigonometric identity makes the computation of the power series for $\cos^2 w$ easy:
$$\cos (2w) = \cos^2 w - \sin^2 w = 2\cos^2 w - 1,$$
and hence $\cos^2 w = \frac{1}{2}(1+\cos (2w))$. Thus
$$\cos^2 \frac{1}{z^2} = \frac{1}{2}\left(1+\cos \frac{2}{z^2}\right) = \frac{1}{2} + \frac{1}{2}\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!} \frac{2^{2n}}{z^{4n}} = 1 + \sum_{n=1}^\infty \frac{(-1)^n2^{2n-1}}{(2n)!z^{4n}}.$$
Multiplying that with $z^6$ yields the result.
In part iv), you need to expand the function in powers of $z-1$ since the domain is an annulus with centre $1$.
$$\frac{1-2z}{z^2-z} = \frac{1}{z-1}\left(\frac{1}{z} - 2\right) = (z-1)^{-1}\left(\frac{1}{(z-1)+1}-2\right).$$
Now expand $\frac{1}{(z-1)+1}$ into a geometric series, keeping in mind that $\lvert z-1\rvert > 1$ in the domain.
In the last, you made a sign error in your partial fraction decomposition,
$$\frac{1}{1+z^2} = \frac{1}{(z-i)(z+i)} = \frac{1}{2i}\left(\frac{1}{z-i} - \frac{1}{z+i}\right).$$
From then on, you have once typed $z$ where it ought to have been $u$, but made no real mistake. So to obtain the correct Laurent series, you only need to adjust the signs to match the correct partial fraction decomposition.