1. Understand the sets

We are given $S=\{a+b\sqrt2:a,b\in\mathbb Z\}.$ Also, $T_1=\{(-1+\sqrt2)^n:n\in\mathbb N\},\qquad T_2=\{(1+\sqrt2)^n:n\in\mathbb N\}.$

We must check each option independently.

2. Useful algebraic fact

Notice that $-1+\sqrt2=\sqrt2-1.$ Also, $(1+\sqrt2)(\sqrt2-1)=1,$ so $\sqrt2-1=\frac1{1+\sqrt2}.$

Further, if a number is of the form $a+b\sqrt2$ with integers $a,b$, then its powers also remain of the same form, because $(a+b\sqrt2)(c+d\sqrt2)=(ac+2bd)+(ad+bc)\sqrt2,$ and the new coefficients are again integers.

Thus powers of $1+\sqrt2$ and $\sqrt2-1$ belong to $S$.

3. Option A: $\mathbb Z\cup T_1\cup T_2\subset S$

• Every integer $m\in\mathbb Z$ can be written as $m=m+0\sqrt2\in S.$ Hence $\mathbb Z\subset S$.

• Since $1+\sqrt2\in S,$ all its natural powers lie in $S$. Hence $T_2\subset S.$

• Since $-1+\sqrt2=\sqrt2-1\in S,$ all its natural powers also lie in $S$. Hence $T_1\subset S.$

Therefore, $\mathbb Z\cup T_1\cup T_2\subset S.$

So A is true.

4. Option B: $T_1\cap\left(0,\frac1{2024}\right)=\phi$

Let $r=\sqrt2-1.$ Then $0<r<1.$ So the sequence $r^n$ is positive and tends to $0$. Hence for sufficiently large $n$, $0<r^n<\frac1{2024}.$ So there do exist elements of $T_1$ inside the interval $\left(0,\frac1{2024}\right)$.

In fact, since $r=\frac1{1+\sqrt2}\approx 0.4142,$ we get $r^9\approx (0.4142)^9\approx 0.00035<\frac1{2024}\approx 0.000494.$ Thus $(-1+\sqrt2)^9\in T_1\cap\left(0,\frac1{2024}\right).$

Therefore the intersection is not empty.

So B is false.

5. Option C: $T_2\cap(2024,\infty)\neq\phi$

Let $s=1+\sqrt2\approx 2.4142>1.$ Then $s^n\to\infty$ as $n\to\infty$. Hence for sufficiently large $n$, $s^n>2024.$ So there exists at least one element of $T_2$ in $(2024,\infty)$.

For example, $(1+\sqrt2)^9\approx (2.4142)^9\approx 1681<2024,$ $(1+\sqrt2)^{10}\approx 4059>2024.$ Thus $(1+\sqrt2)^{10}\in T_2\cap(2024,\infty).$

So C is true.

6. Option D: For any $a,b\in\mathbb Z$, $\cos(\pi(a+b\sqrt2))+i\sin(\pi(a+b\sqrt2))\in\mathbb Z\iff b=0.$

Let $z=\cos(\pi(a+b\sqrt2))+i\sin(\pi(a+b\sqrt2))=e^{i\pi(a+b\sqrt2)}.$

We need to understand when $z\in\mathbb Z$.

Since $|z|=1,$ any integer value of $z$ must be an integer of modulus $1$. The only such integers are $z=1\quad\text{or}\quad z=-1.$

So $z\in\mathbb Z$ iff $e^{i\pi(a+b\sqrt2)}=\pm 1,$ which happens iff $a+b\sqrt2\in\mathbb Z.$ Because $a\in\mathbb Z$, this requires $b\sqrt2\in\mathbb Z.$ But $b\in\mathbb Z$ and $\sqrt2$ is irrational, so this is possible iff $b=0.$

Conversely, if $b=0$, then $z=\cos(\pi a)+i\sin(\pi a)=(-1)^a\in\{1,-1\}\subset\mathbb Z.$

Hence $\cos(\pi(a+b\sqrt2))+i\sin(\pi(a+b\sqrt2))\in\mathbb Z \iff b=0.$

So D is true.

7. Final conclusion

The true statements are: $\boxed{A,\ C,\ D}$

This matches the stored correct answer.