Is there a model $\mathcal{M}$ of Th($\mathbb{N}$) in the language $L=\{0,1,+,\cdot\}$ such that there is an element $m\in \mathcal{M}$ with $0This is exercise 8.1 from the book "An Invitation to Model Theory" by J. Kirby. I have no clue how to approach this problem.
Also, another question that might be kinda stupid is:
Let $\mathcal{M}$ be an $L-$structure for some first-order language $L$ and $R$ be a non-definable relation on $\mathbb{M}$.
Let Th$(\mathcal{M})=K$ be the set of all $L-$sentences $\mathcal{M}$ satisfies.
Let $L'=L\cup{R}$ where R is a relation symbol. Then $\mathcal{M}$ is an $L'-structure$ with interpretation of $R$ is the relation $R$ on $M$.
Can we say that there is a formula $\phi$ which contains the symbol $R$ such that $\phi$ is independent of $K$?
 A: Addressing your follow-up question; the answer is absolutely yes, and you don't even need the relation to be undefinable! For some minor convenience, I'm going to assume the underlying set $M$ of the $L$-structure $\mathcal{M}$ is infinite.
Note that $K$ says nothing about the symbol $R$; $K$ is an $L$-theory, and the symbol $R$ is not contained in $L$ by hypothesis. There are in particular infinitely many ways to expand $\mathcal{M}$ into an $L\cup \{R\}$-structure – one for every subset $A\subseteq M^n$, where $n$ is the arity of $R$. Even though you started with the question with a particular choice of relation on $M$ in mind, there's absolutely no reason you have to stick with it when interpreting a new symbol.
Thus, for example, you can take $\phi\equiv\forall x_1\dots\forall x_n R(x_1,\dots,x_n)$, or you can take $\phi\equiv\neg\exists x_1\dots\exists x_n R(x_1,\dots,x_n)$, and in either case $\phi$ will be independent of $K$.
In fact, it is the case that $\mathcal{M}$ can be turned into a model of any $R$-sentence $\phi$ with an infinite model. (Hence, if $\phi$ is any $R$-sentence such that both $\phi$ and $\neg\phi$ have infinite models, then $\phi$ will be independent of $K$.) This is a special case of the following general fact:
Theorem: Let $L_1$ and $L_2$ be countable disjoint languages, and let $T_1$ and $T_2$ be $L_1$ and $L_2$ theories each having an infinite model. Then $T_1\cup T_2$ is consistent, and in fact any infinite model of $T_1$ may be made into a model of $T_{2}$.
Proof: Fix any infinite model $\mathcal{M}\models T_1$, with underlying set $M$. By Löwenheim-Skolem, $T_{2}$ has a model $\mathcal{N}$, with underlying set $N$, such that $|N|=|M|$. Choose any bijection $\alpha:N\to M$.
We will define an $L_1\cup L_2$-structure $\mathcal{M}'$, with underlying set $M$, as follows. For any function symbol, constant symbol, or relation symbol $\mathfrak{g}\in L_1$, let $\mathfrak{g}^{\mathcal{M}'}=\mathfrak{g}^{\mathcal{M}}$. This handles the symbols in $L_1$.
For the symbols in $L_2$, we use "transport of structure" along $\alpha$. For a constant symbol $c\in L_{2}$, we take $c^{\mathcal{M}'}=\alpha(c^{\mathcal{N}})$. For a relation symbol $R\in L_2$ we take $R^{\mathcal{M}'}=\alpha(R^{\mathcal{N}})$. Finally, for a function symbol $f\in L_2$ we take $f^{\mathcal{M}'}(\overline{m})=\alpha(f^{\mathcal{N}}(\alpha^{-1}(\overline{m})))$ for every $\overline{m}$ a tuple from $M$ of the same arity as $f$.
Since $L_1$ and $L_2$ are disjoint, the process indicated above is well-defined, and makes $\mathcal{M}'$ into an $L_1\cup L_2$-structure with underlying set $M$.
Exercise: Show that the $\alpha$ is an isomorphism of $L_2$-structures between $\mathcal{N}$ and the reduct (are you familiar with this term?) of $\mathcal{M}'$ to $L_2$. $\square$
In particular, $\mathcal{M}'$ is a model of $T_2$. Also, the reduct of $\mathcal{M}'$ to $L_1$ is equal to $\mathcal{M}$, and hence by hypothesis a model of $T_1$. So $\mathcal{M}'\models T_1\cup T_2$, as desired. $\blacksquare$

Now, all of that said, this is not the intention of the exercise. You are right that the symbol '$<$' is not contained in $L=\{0,1,+,\cdot\}$, and so the question as written is ambiguous. But implicit in the exercise is that the standard ordering on $\mathbb{N}$ is definable in $(\mathbb{N},0,1,+,\cdot)$, by the formula $\phi(x,y)\equiv\exists z(y=x+z)$, and so the symbol $<$ in the question is meant to be identified with this first-order definition. In other words, the exercise wants you to take the symbol $<$ as a shorthand for the formula $\phi(x,y)$. Can you now see how to do the question? A hint is written below.

 Hint: Note that $(\mathbb{N},0,1,+,\cdot)\models\neg\exists x(\phi(0,x)\wedge\phi(x,1))$.

