The example is given below:
My Questions are:
1- In the first line, why "then $\vartriangle_{0} (S^1)$ and $\vartriangle_{1} (S^1)$ are both $\mathbb{Z}$" ?
2-I do not understand the last paragraph in the given picture. could anyone explain it for me, please?
3-Is the homology group of the circle equal to its simplicial homology group?
To answer $1$, you'll find earlier in the chapter that $\Delta_n(X)$ is defined to be the free abelian group with a generator for every $n$-simplex in $X$.
In this example, the number of $0$-simplices (= vertices) is $1$, because there is just one vertex $v$. So $\Delta_0(X)$ is the free abelian group on 1 generator (and every free abelian group on $1$ generator is isomorphic to $\mathbb Z$), and he is calling that 1 generator $v$.
Also, the number of $1$-simplices (= edges) is $1$ because there is just one edge $e$. So again $\Delta_1(X)$ is the free abelian group on $1$ generator, which is isomorhpic to $\mathbb Z$, and he is calling that $1$ generator $e$.
To answer $2$, he has computed $\partial_1 e = v - v = 0$. Therefore, $\partial_1 : \Delta_1(X) \mapsto \Delta_0(X)$ takes the generator $e$ to zero. Therefore $\partial_1$ is the zero homomorphism. It's even easier to see that every other boundary map in this chain complex is also the zero homomorphism because either its domain or its range is the trivial group.
Now I'll leave you to work it out: what's the kernel and the range of a zero homomorphism? How do you use that to compute the homology groups, when all the boundary maps are zero homomorphisms?
Your question 3 is not well formulated. I expect you did not intend to ask "Is the simplicial homology group of a circle equal to its simplicial homology group", but then I would wonder what kind of homology group you are really asking about? There is a more advanced theorem that Hatcher is working up to in that portion of the book, which says that the singular homology group of any $\Delta$ complex is isomorphic to (not equal to) its simplicial homology group.