It appears to be that the x-Interval (0,1) has been divided into sub intervals $(c_{i-1},c_i)$ for $i=1,\ldots,n$.
a) The $c_i$ are defined in such a way as to make $f(c_i)$ easy to evaluate (to $\frac{i}{n}$), probably in order to make the evaluation of the limit $$\lim\limits_{n \to ∞} \sum_{i=1}^n f(c_i)\Delta x_i$$ not so technically demanding.
b) $\Delta x_i$ as defined is the length of the above mentioned interval $(c_{i-1},c_i)$. That makes sense as Rieman Sums of a function are sums of products of interval length ($\Delta x_i$) and a function value from that interval ($f(c_i)$).
Answering questions from OP:
Please use the below answer from Paramanand Singh if you need to understand what the Rieman Sums are. I assumed above that you basically know them, but are unsure how they apply to this problem.
The values $\frac{i^2}{n^2}$ are just a convenient way to partition the interval [0,1]. They come from "having solved lots of similar problems and thus having a repertoire of tools for them" on the part of the textbook author. You are, I think, not supposed yet to come up with them yourself, you are just supposed to follow along and solve the problem as stated. Later when learning about integrals you will see why this approach was chosen, what alternatives there were, etc.
The length given for $\Delta x_i$ is just the normal formula for the length of an interval: upper boundary - lower boundary. Since the upper boundary is $c_i=\frac{i^2}{n^2}$ and the lower boundary is $c_{i-1}=\frac{(i-1)^2}{n^2}$ you get the stated formula for $\Delta x_i$
The right Riemann sum using the partition with points $x_k = 2 \sin \frac{\pi k}{2n}\,\,(k = 0,1,2,\ldots, n)$ is
$$\tag{*}S_n = \sum_{k=1}^n \sqrt{4 - 4 \sin^2 \frac{\pi k}{2n}}\left(2 \sin \frac{\pi k}{2n}- 2\sin \frac{\pi (k-1)}{2n}\right)\\ = 4\sum_{k=1}^n \cos \frac{\pi k}{2n}\left( \sin \frac{\pi k}{2n}- \sin \frac{\pi k}{2n}\cos \frac{\pi}{2n}+ \sin \frac{\pi }{2n}\cos \frac{\pi k}{2n}\right)\\ = 4\left(1 -\cos\frac{\pi}{2n} \right)\sum_{k=1}^n \cos \frac{\pi k}{2n}\sin \frac{\pi k}{2n} + 4 \sin \frac{\pi}{2n}\sum_{k=1}^n \cos^2 \frac{\pi k}{2n}$$
Note that
$$\sum_{k=1}^n \cos \frac{\pi k}{2n}\sin \frac{\pi k}{2n}= \frac{1}{2}\sum_{k=1}^n\sin \frac{2\pi k}{2n} =\frac{1}{2}\sum_{k=1}^n\sin \frac{\pi k}{n},\\\sum_{k=1}^n \cos^2 \frac{\pi k}{2n}= \sum_{k=1}^n \left(\frac{1}{2} - \frac{1}{2} \cos \frac{2\pi k}{2n} \right) = \frac{n}{2}- \frac{1}{2}\sum_{k=1}^n\cos \frac{\pi k}{n}$$
Substituting into (*), we get
$$S_n = \underbrace{2\left(1 -\cos\frac{\pi}{2n} \right)\sum_{k=1}^n \sin \frac{\pi k}{n}}_{A_n} + \underbrace{2n \sin \frac{\pi}{2n}}_{B_n}- \underbrace{\sin \frac{\pi}{2n}\sum_{k=1}^n \cos\frac{\pi k}{n}}_{C_n} $$
It is shown below that $A_n - C_n \to 0$ as $n \to \infty$, and, therefore,
$$\int_0^2 \sqrt{4-x^2}\, dx= \lim_{n \to \infty}S_n = \lim_{n \to \infty}2n \sin \frac{\pi}{2n} = \lim_{n \to \infty}\pi \frac{\sin \frac{\pi}{2n}}{\frac{\pi}{2n}} = \pi$$
Limits of $A_n$ and $C_n$:
We have, since $\frac{1-\cos x}{\sin x} \to 0$ as $x \to 0$,
$$A_n = 2\left(1 -\cos\frac{\pi}{2n} \right)\sum_{k=1}^n \sin \frac{\pi k}{n}= 2 \left(1 -\cos\frac{\pi}{2n} \right) \frac{\sin \frac{n}{2}\frac{\pi}{n}\sin \frac{n+1}{2} \frac{\pi}{n}}{\sin \frac{\pi}{2n}}\\ = 2 \frac{1 -\cos\frac{\pi}{2n} }{\sin \frac{\pi}{2n}}\sin \left[\left(1 + \frac{1}{n}\right)\frac{\pi}{2}\right]\underset{n\to \infty}\longrightarrow 0$$
Also,
$$C_n = \sin\frac{\pi}{2n} \sum_{k=1}^n \cos \frac{\pi k}{n} = \sin\frac{\pi}{2n}\frac{\sin \frac{n}{2}\frac{\pi}{n}\cos \frac{n+1}{2} \frac{\pi}{n}}{\sin \frac{\pi}{2n}}\\ = \cos \left[\left(1+\frac{1}{n} \right)\frac{\pi}{2}\right]\underset{n\to \infty}\longrightarrow 0$$
Best Answer
I see two potential approaches:
Try computing both integrals between $[0, 2]$ and $[0, 1)$, and subtract afterwards.
If you want to stick to a partition of $[1, 2]$ you could still make the partition follow the same pattern. Try $\frac{i^2}{n^2}$ for $i$ going from $n$ to $\sqrt{2}n$, for example.