As given here, let $n\ge 3$ be an integer. Also, the complex number $z_n= \cos(2\pi/n) + i\sin(2\pi/n)$ has the modulus $1,$ and is at an angle $\theta_n= 2\pi/n$ wrt the positive real axis.
The set $C_n=\langle z_n\rangle= \{1, z_n, z_n^2, \cdots, z_n^{n-1}\}$ is a cyclic group of order $n,$ under the operation of composition.
Have use of the DeMoivre's theorem to find the different terms, under the group operation of composition, as given by: $(\cos(\theta) + i\sin(\theta))^k = \cos(k\theta) + i\sin(k\theta),$ for $k=1,2,\cdots$
So, for $n=3,$ should have $C_3=\langle z_3\rangle$ $ = \{ 1, (\cos(\pi/3) + i\sin(\pi/3)), (\cos(2\pi/3) + i\sin(2\pi/3))^2) \}.$
Q. If $(z_n^i)^{-1}= z_n^j,$ for some $j$ with $0\le j\lt n,$ what is $j$ in terms of $i?$
The answer seems: $j= n-i.$
Have another doubt, as why in the first line $n$ cannot have value $=2.$
There are some things incorrect here, and some things missing context. Here is what's incorrect: the group of $3$rd roots of unity should be: $$ C_3 = \{ 1, e^{2\pi i/3}, e^{4\pi i/3}\} $$ which of course you can put in terms of sines and cosines using Euler's formula (as you've written, the arguments are incorrect, or you're missing powers). To be clear, in your notation, $z_3 = e^{2\pi i/3}$.
Also, this is not exactly a mistake, but I would warn against using $i$ as an index when speaking of complex numbers.
Now, for your first question: observe that $z_n^j * z_n^{n-j} = z_n^{n} = 1$, so you are correct, $z_n^{n-j}$ is the inverse of $z_n^j$. In the case that $j \geq n$, you reduce $j$ modulo $n$ first (since having $n - j$ be non-negative is probably helpful to think about).
For the last question, $n \geq 3$ is just an assertion that makes sense only if you read the preceding paragraph in the picture you sent (Question 4.13.2). That paragraph effectively addresses the $n = 2$ case (since the $2$nd roots of unit are just $1$ and $-1$). So since we are generalizing that case, we "now consider $n \geq 3$".