1) How can I prove that $\mathbb Z[x]/(1+x^2)\mathbb Z[x]$ is a free $\mathbb{Z}$-module with basis $\{1,\bar x\}$?
I wanted to prove that $$\mathbb Z[x]/(1+x^2)\mathbb Z[x]\cong \mathbb Z^2,$$ but it looks complicate.
2) Is $\mathbb Z[x]$ a free $\mathbb Z-$module? I would say yes and that $\{1,x,x^2,...\}$ is a basis, but how can I prove it?
1) Over any commutative ring $R$, the quotient ring $R[X]/(f(X))$ of $R[X]$ by a monic polynomial is a finitely generated free $R$-module, with rank equal to the degree of the polynomial.
Denoting by $x$ the class of $X$ in the quotient, you just have to prove that any $x^n$, with $n\ge \deg f$ lies in the submodule generated by $\;1, x,\dots,x^{\deg f}$ (simple induction), and that these elements are linearly independent.
2) You're perfectly right. It is part of the definition: $R[X]$ is the free algebra of the monoid $\mathbf N$. In particular, as an $R$-module it is simply $R^{(\mathbf N)}$ (functions from $\mathbf N$ to $R$ with finite support).
Re 1: If you want to follow through with your method, then start by showing that each class modulo $(1+x^2)$ contains a unique element of degree at most $1$. To this end you might for example use euclidean division (note that you can always divide by $1 + x^2$ as its leading coefficient, $1$, is invertible in the integers).
Having done that you can define directly a map from $\mathbb{Z}[x]/(1+x^2)\mathbb{Z}[x]$ to $\mathbb{Z}^2$ by $\overline{P} \mapsto (a,b)$ where $a+bx$ is the polynomial of degree at most $1$ in the class $\overline{P}$. (This is well defined once you have shown that there is a unique such polynomial for each class.)
It is then easy to argue that the map is bijective. (Injective is essentially already done by the uniqueness argument, and surjective is quite direct.)
It remains to show that it is a module homomorphism, that is, the map is compatible with the module operations, but that is not that hard to check either.
Re 2: Yes, that's true. How to prove it depends a bit on you definition of the polynomial ring $\mathbb{Z}[x]$. But given a polynomial $a_0+a_1x + a_2x^2 + \dots + a_nx^n$ you directly see that it is a linear combination of the $1, x, x^2, \dots$. As it is $a_0 \times 1 + a_1 \times X + a_2 \times x^2 + \dots + a_n \times x^n$. It remains to show that $1, x, x^2, \dots$ is linearly independent.
Usually, I would say this is immediate by the definition of a polynomial as a polynomial is uniquely identified by the sequence of its coefficients, and thus the only way to get the zero-polynoimal is as $0 \times 1 + 0 \times x + 0 \times x^2 + \dots$.
If you want a more of a proof there you could say that given whatever $a_0 + \dots a_nx^n$ with some $a_i$ non-zero you can find some integer where its evaluation is non-zero, as it can have at most $n$ roots, contra the assumption it equal the zero-polynomial.