The task asks me to calculate the curve integral of that function $F(z)$ over the curve $\gamma(t) = \{z \in\mathbb{C}: \:\vert z\vert=1\}$. Apparently this is a circle in the complex plane.
Before I proceed to my pure calculation process I actually wonder what that calculation will mean. If I were to look closer at $F(z)$ I should notice how it maps a function from 2 Dimensions to 4 Dimensions (each plugged in complex number is matched with another complex number, both depending on 2 variables). Now how does this plot touch the curve $\gamma$? Is this similar to ordinary curve integrals, where all Function values of the plot along the curve are added together?
Anyway, here my calculations, that I came up with just by intuition:
$$\begin{align} &\text{parametrisation of $\gamma(t)$ by}\: e^{i\,t}\,\quad t\in[0,2\,\pi] \\\\ &\text{ into the usual definitionn of curve integral:}\\\\ &\int_C F(z)\,\mathrm{dz} = \int_0^{2\,\pi} F(\gamma(t))\,\gamma'(t)\,\mathrm{dt} = \\\\ & \int_0^{2\,\pi}e^{i\,t\,(2-k)}\,i\,e^{i\,t}\,\mathrm{dt} = \left[\dfrac{1\,i}{i\,(3-k)}\,e^{i\,t(3-k)}\right]_0^{2\,\pi} = \\\\ &\dfrac{1}{(3-k)}\,\left[\cos\bigr((3-k)\,2\,\pi\bigl)+i\sin\bigr((3-k)\,2\,\pi\bigl)-1\right] = \\\\ &\dfrac{1}{3-k}\left(1-1\right) = 0 \quad \text{is this the answer?} \end{align}$$
Edit:
coming back to this I still need help for $k = 3$, that peculiar case.
I thought it'd be easy to show the integrals also equals $0$, instead:
$\displaystyle{\int_{0}^{2\,\pi}e^{i\,t(3-3)}\,i\,\mathrm{dt} = \left[i\,t\right]_0^{2\,\pi}} = 2\,\pi\,i$ ?
Also having access to the Residue Theorem now it seems to agree: $\displaystyle{\int_{C}F(z)}=\displaystyle{\int_{C}\dfrac{z^2}{z^k}} = (\text{Res$(0)$} + \cdots + \text{Res$(0)$})\,2\,\pi\,i$ for each k from $1 \cdots n$
hence $\displaystyle{\int_{C}F(z) = (\lim_{z \to 0}\dfrac{z^2}{z^1}\,z^1 + \lim_{z \to 0}\dfrac{d}{dz}\dfrac{z^2}{z^2}\,z^2\,\dfrac{1}{1!}+ \lim_{z \to 0}\dfrac{d^2}{dz^2}\dfrac{z^2}{z^3}\,z^3\,\dfrac{1}{2!}+\cdots)\,2\,\pi\,i} $
Here likewise for $k = 3$ the integral seems to become non zero: $\int_C F(z) =2\,\pi\,i$. The rest vanishes. It just happens to disagree with taking the limit of the very first expression: $$\displaystyle{\lim_{k\to 3}\dfrac{1}{3-k}\left(1-1\right) = 0 ?}$$
Most of what you have done is correct, but there are two issues. First you seem to have a misconception of what the residue theorem says:
This is not correct. There is only one residue at any point. Even if you have a $N$th order pole at some point then you don't sum up $N$ residues - only one! For your function $F(z) = z^{2-k}$ the only pole is $z=0$ and the residue here (you can read it straight off the expression as its already in the form of a Laurent series about $z=0$) is $0$ if $k\not=3$ and $1$ if $k=3$ so $$\displaystyle{\int_{C}F(z)}=\displaystyle{\int_{C}\dfrac{z^2}{z^k}} = 2\pi i\,\text{Res}(F(z),z=0)=\left\{\matrix{0, & k\not=3\\2\pi i, & k=3}\right.$$
You anyway happen to get the correct answer here.
The second issue is below "It just happens to disagree with taking the limit of the very first expression". Here you have to be a bit more careful when evaluating the limit. You seem to just have looked at your expression and noted that the numerator is zero so therefore the limit must be zero, but when $k=3$ you have a $0/0$ expression. Doing it more carefully we see that $$\lim_{k\to 3} \frac{\cos(2\pi(3-k)) + i\sin(2\pi(3-k)) - 1}{3-k} = \lim_{x\to 0} 2\pi\left[\frac{\cos(x)-1}{x} + i\frac{\sin(x)}{x}\right] =2\pi i$$ where I have put $x = 2\pi(3-k)$ and used some well known trigonometrical limits: $\sin(x)\sim x$ and $\cos(x) \sim 1 - x^2/6$ as $x\to 0$. This agrees perfectly with the other result you have found so all is nice and consistent.