Let $S=\{a+b\sqrt{2}:a,b\in <!--\; \in \; -->\mathbb{Z}\},T_1$$=\{(-<!--\; -\; -->1+\sqrt{2}{)}^n:n\in <!--\; \in \; -->\mathbb{N}\}$and $T_2=\{(1+\sqrt{2}{)}^n:n\in <!--\; \in \; -->\mathbb{N}\}$. Then which of the following statements is (are) TRUE?

To determine which of the statements are true, let us analyze each of them step-by-step.

First, recall the definitions of the sets given in the problem statement:
$S=\{a+b\sqrt{2}:a,b\in \mathbb{Z}\}$
$T_1=\{(-1+\sqrt{2}{)}^n:n\in \mathbb{N}\}$
$T_2=\{(1+\sqrt{2}{)}^n:n\in \mathbb{N}\}$

Step 1: Analyze the statement $\mathbb{Z}\cup T_1\cup T_2\subset S$
Let us check if each of the subsets $\mathbb{Z}$, $T_1$, and $T_2$ is contained in $S$.
1. For any integer $k\in \mathbb{Z}$, we can write $k=k+0\sqrt{2}$. Since $k,0\in \mathbb{Z}$, we have $k\in S$. Thus, $\mathbb{Z}\subset S$.
2. Note that the set $S$ is closed under addition, subtraction, and multiplication because it is the ring of integers of the quadratic field $\mathbb{Q}(\sqrt{2})$. Specifically, if $x,y\in S$, then $x\cdot y\in S$.
Since $-1+\sqrt{2}\in S$ (with $a=-1,b=1$), any positive integer power $(-1+\sqrt{2}{)}^n$ must also belong to $S$ by mathematical induction. Hence, $T_1\subset S$.
3. Similarly, since $1+\sqrt{2}\in S$ (with $a=1,b=1$), any positive integer power $(1+\sqrt{2}{)}^n$ must also belong to $S$. Hence, $T_2\subset S$.
Combining these results, we get $\mathbb{Z}\cup T_1\cup T_2\subset S$. Therefore, this statement is TRUE.

Step 2: Analyze the statement $T_1\cap (0,\frac{1}{2024})=\varphi$
Let us analyze the base of the exponent in $T_1$:
Since $1.414<\sqrt{2}<1.415$, we have:
$0<-1+\sqrt{2}<0.415<1$
For a base $r=\sqrt{2}-1$ where $0<r<1$, the sequence $x_n=r^n$ is strictly decreasing and approaches $0$ as $n\to \mathrm{\infty }$.
Thus, for a sufficiently large $n$, we can always find $r^n<\frac{1}{2024}$.
Therefore, the intersection $T_1\cap (0,\frac{1}{2024})$ is not empty. This statement is FALSE.

Step 3: Analyze the statement $T_2\cap (2024,\mathrm{\infty })\ne \varphi$
Let us analyze the base of the exponent in $T_2$:
Since $1+\sqrt{2}\approx 2.414>1$, the sequence $y_n=(1+\sqrt{2}{)}^n$ is strictly increasing and diverges to $\mathrm{\infty }$ as $n\to \mathrm{\infty }$.
Thus, we can definitely find a positive integer $n$ such that $(1+\sqrt{2}{)}^n>2024$ (for instance, ${2.4}^9>2500>2024$).
Therefore, the intersection $T_2\cap (2024,\mathrm{\infty })$ is non-empty. This statement is TRUE.

Step 4: Analyze the statement: For any given $a,b\in \mathbb{Z}$, $\cos (\pi (a+b\sqrt{2}))+i\sin (\pi (a+b\sqrt{2}))\in \mathbb{Z}$ if and only if $b=0$
Using Euler's formula, let:
$z=\cos (\pi (a+b\sqrt{2}))+i\sin (\pi (a+b\sqrt{2}))=e^{i\pi (a+b\sqrt{2})}$
If $z\in \mathbb{Z}$, then $z$ must be a real number. For $z$ to be real, the imaginary part must be zero:
$\sin (\pi (a+b\sqrt{2}))=0$
This occurs if and only if the argument is an integer multiple of $\pi$:
$\pi (a+b\sqrt{2})=k\pi \text{for some }k\in \mathbb{Z}$
Dividing by $\pi$:
$a+b\sqrt{2}=k$
Rearranging the equation:
$b\sqrt{2}=k-a$
Since $a,k\in \mathbb{Z}$, the right-hand side $k-a$ is an integer.
If $b\ne 0$, then $\sqrt{2}=\frac{k-a}{b}$, which implies that $\sqrt{2}$ is rational. This is a contradiction since $\sqrt{2}$ is irrational.
Thus, we must have $b=0$.
Conversely, if $b=0$, then:
$z=\cos (\pi a)+i\sin (\pi a)=(-1{)}^a$
Since $a\in \mathbb{Z}$, $(-1{)}^a\in \{-1,1\}\subset \mathbb{Z}$.
Hence, $z\in \mathbb{Z}$ if and only if $b=0$. This statement is TRUE.