Let $D$ be integral domain. We know that we are able to construct the field of quotients $F$. More exactly the field $F$ consists of all equivalence classes $[a,b]$, where $a\in D$, $b\in D-\{0\}$ and
$$[a,b]=\{(c,d)\in D\times D-\{0\}: (a,b)\sim(c,d)\},$$
where we claim that $(a,b)\sim (c,d)$ iff $ad=bc$.
It is obvious that $D$ is not the subset of $F$, so we cannot say that $D\subset F$.
The correct statement is the following: $D \hookrightarrow F$ i.e. there exists $\phi: D\to F$ such that $\phi$ - injective homomorphism.
Let's move on to the questions.
1) We know that $\mathbb{Z}=\{0,\pm 1,\pm 2,\dots\}$ the set of integers and $\mathbb{Q}$ the set of rational numbers.
However $\mathbb{Q}$ is the set of all equivalence classes $[a,b]$, where we can denote $[a,b]$ by symbol $\frac{a}{b}$. The set of integers $\mathbb{Z}$ can be embedded into $\mathbb{Q}$ through mapping $\phi:\mathbb{Z}\to \mathbb{Q}$ defined by $\phi(n)=[n,1]=\frac{n}{1}$.
We cannot say that $\mathbb{Z}\subset \mathbb{Q}$ because their elements have distinct nature. But why many authors write that $\mathbb{Z}\subset \mathbb{Q}$?
2) My second question also deals with this topic so I've decided not to create new one.
Let $R$ be Euclidean ring. We know that in Euclidean ring $R$ greatest common divisor and least common multiple always exist! And my book states that $\text{lcm}(a,b)$ and $\text{gcd}(a,b)$ have the following relation: $$\text{lcm}(a,b)=\dfrac{ab}{\text{gcd}(a,b)}.$$
I have big misunderstanding with this question. The LHS of this identity is the $\text{lcm}(a,b)$ which is an element of Euclidean ring $R$ but the RHS of identity is the fraction $\frac{ab}{\text{gcd}(a,b)}$ which is an equivalence class and it is the element of quotient field but not an element of Euclidean ring $R$. There is such a question: How can we say that they are equal? if they even have different nature.
I would be very very thankful for detailed explanation since I am trying to understand both question about 3 weeks but no results. Also I was trying to find some useful links here but it did not give results. So please help me!
The answer is simple: we identify $D$ (or $\mathbf Z$) with its canonical image in $F$ ($\mathbf Q$).
Recall that $\mathbf Z$ itself is constructed from $\mathbf N$ as the set of equivalence classes in $\mathbf N^2$ for the relation $$(a,b)\sim (c,d)\overset{\text{def}}{\iff} a+d=b+c,$$ so $\mathbf N$ isn't really a subset of $\mathbf Z$, yet we use the identification of $\mathbf N$ with its canonical image in $\mathbf N^2/\sim$.
The answer to your second question follows from these considerations. Note that $\;\dfrac{ab}{\gcd(a,b)}$ can be interpreted a the quotient of $ab$ by $\gcd(a,b)$, so it's an element of the Euclidean domain. And anyway the relation can be written entirely in $D$ as $$\operatorname{lcm}(a,b)\,\gcd(a,b)=ab.$$