I need to find the integral over smooth curve $C$ of
1) $\displaystyle \frac{e^{iz}}{z^2}\, dz$ when $r(t) =e^{it}, t\in [0, 2\pi]$
2) $\dfrac{\log z}{z^n} \, dz$, $r(t) = 1 +(1/2) e^{it}, t\in [0,2\pi]$ & $n\ge0$.
I know that I can use $e^{it} = \cos t + i\sin t$ and find the integral. But things are getting messy for me. Please help.
Assuming you don't know Cauchy's integral theorem / residue theorem, we will still do something close.
$$e^{iz} = \sum \frac {(iz)^n}{n!}$$
$$\int \frac {e^{it}}{z^2} \, dz= \int \frac 1{z^2} \, dz + \int \frac i z \, dz + \int p(z) \,dz$$ where $p(z)$ is a polynomial
$$\int \frac 1{z^2}dz = \int_0^{2\pi} i e^{-it} \;dt = - e^{-it}|_0^{2\pi} = 0$$
$$\int \frac i z \, dz = \int_0^{2\pi} -1\; dt = -2\pi$$
$$\int p(z) \, dz = 0$$
I am going to leave it to you to prove that last one to yourself if you haven't learned it already.
Can you do a similar expansion on $\ln z$?
Your first integral is $$ \int_C \frac{e^{iz}}{z^2} \,dz $$ where $C$ is the circle of radius $1$ centered at $0$. The function $z\mapsto e^{iz}$ is differentiable everywhere in $\mathbb C$ and $z\mapsto1/z^2$ is differentiable everywhere in $\mathbb C$ except at $0.$ The quotient approaches $\infty$ at $0.$ Therefore the integral is the same for every curve $C$ that winds once counterclockwise about $0.$ As the radius of $C$ shrinks, the numerator become indistinguishable from the function everywhere equal to $1$, and the integral approaches $\displaystyle \int_C \frac 1 {z^2} \,dz.$ This integral also does not depend on the radius, for the same reason. So see if you can figure out what that is.
For the second integral, note that $$ \log z = (z-1) + \sum_{n=2}^\infty c_n (z-1)^n $$ and $z\mapsto 1/z^n$ is differentiable everywhere except at a point that is outside the curve $C$. Therefore the integral is $0.$