$Qustions$
Suppose $\mathcal{M_n} = (M_n ,I_n )$ is an $L$-model, $\mathcal{M_n}$$\preceq$$\mathcal{M_{n+1}}$ for each $n∈\mathbb N$. Let$M = \bigcup_{n \in \mathbb N} M_n $
1.Define an interpretation $I$ on $M$ properly so that each $\mathcal{M_n}$ is a substructure of $\mathcal{M} = (M,I)$.
2.Show $\mathcal{M_n}$$\preceq$$\mathcal{M}$ for each $n$ and the above $\mathcal{M}$.
3.Show if $\mathcal{N}$ is such that $\mathcal{M_n}$$\prec$$\mathcal{N}$ for all $n$, then $\mathcal{M}$$\preceq$$\mathcal{N}$.
$Answer$
First part
I have finished the first part.
1.For constant symbol $c$, $I(c)=I_0(c)$
2.If $F$ is an $n$-ary function symbol, $a_1,...,a_{n+1}∈M_i$, then $I(F)(a_1, \ldots, a_n) =a_{n+1}$ iff$ I(F)(a_1, \ldots, a_n)=$$a_{n+1}$.
3.If $R$ is a $n$-ary relation symbol, $a_1,...,a_n∈M_i$, then$(a_1, \ldots, a_n)∈I(R)$ iff$ (a_1, \ldots, a_n)∈$$I(R_i)$.
The second part
I think we need $Tarski's$ $Criterion$ in this part. Suppose $A$ is definable in $\mathcal{M}$ with parameters from $M_n$, then we need to show $A∩M_n≠∅$. But the difficult is how to find a element which belongs to both $A$ and $M_n$?
the third part
Intuitively, it is obvious. But how to prove? Still use $Tarski's Criterion$?
Summary
I hope you can give some help in $Question 2,3$
Remember that $M$ is the union.
Hence if you have a witness in $M$, then there must be a witness in some $M_{k}$ for some $k$. Now using the fact that $M_{n}\leq M_{k}$
I'm not sure about your question 3, unless you mean "then $M\leq N$". In which case once again you just need to check the definition: certainly $M$ is a substructure of $N$, and if $N$ have a solution then $M_{n}$ have a solution, hence $M$.