For the first part:
The polynomial ring $\mathbb{Z}_2[x]$ has a universal property:
Given any commutative ring with unity $S$, any ring homomorphism $g\colon\mathbb{Z}_2\to S$, and any $s\in S$, there exists a unique homomorphism $f\colon \mathbb{Z}_2[x]\to S$ such that $f(x)=s$ and $f(a)=g(a)$ for all $a\in\mathbb{Z}_2$.
Here, we start with the ring $S=\mathbb{Z}_2[x]$, $g$ the embedding of $\mathbb{Z}_2\hookrightarrow S$, and take the element $x+1$. So we have a unique morphism $f\colon\mathbb{Z}_2[x]\to\mathbb{Z}_2[x]$ that sends $x$ to $x+1$.
Now compose $f$ with the quotient map $\mathbb{Z}_2[x]\to \mathbb{Z}_2[x]/\langle x^3+x^2+1\rangle$. Call this composite map
$$g\colon \mathbb{Z}_2[x]\to \frac{\mathbb{Z}_2[x]}{\langle x^3+x^2+1\rangle}.$$
The map is surjective, because both $f$ and the projection are surjective. What is its kernel? The kernel is generated by the pre-image of $x^3+x^2+1$: but this is precisely $x^3+x+1$, because as you note $(x+1)^3 +(x+1) + 1 = x^3+x^2+1$. Thus, by the Isomorphism Theorems,
$$\frac{\mathbb{Z}_2[x]}{\langle x^3+x^2+1\rangle} \cong \frac{\mathbb{Z}_2[x]}{\mathrm{ker}(g)} = \frac{\mathbb{Z}_[x]}{\langle x^3+x+1\rangle},$$
proving the desired isomorphism.
For the second part: we use the fact that $(x^2-1) = (x^2+1) = (x+1)^2$. Once again, take the map that sends $x\to x+1$ to get a morphism $f\colon\mathbb{Z}_2[x]\to \mathbb{Z}_2[x]/\langle (x+1)^2\rangle$. Prove the map is surjective, with kernel $\langle x^2\rangle$.
Note this is not going to work over a field that is not of characteristic two, since then $\langle x^2\rangle$ is the square of an ideal, while $\langle x^2-1\rangle = \langle x+1\rangle\langle x-1\rangle$ is the product of two coprime ideals. So the left side would have nilpotent elements, whereas by the Chinese Remainder Theorem the right hand side would be isomorphic to $F\times F$, which does not.
