Which are integral domains? Fields?

For (a), an element of a ring being reducible doesn't mean that it generates a maximal ideal, so you can't conclude that you get a field. (Additionally, it is false that a polynomial with no zero in $\mathbb{Z}$ is irreducible over $\mathbb{Z}$. For instance, consider $4x^2 - 1$.)

Your solution to (b) suffers from the same problem, although in this case you can rescue it by using a property of $\mathbb{F}_5[x]$ that $\mathbb{Z}[x]$ does not have.

For (c), note that no polynomial of degree higher than 2 can be irreducible over the reals, as noted in the comments.

(a) In a UFD (like $\mathbb Z[x]$) the irreducible elements are prime (see here), and prime elements generate prime ideals. In your case, $p(x)=x^2+2x+3$ is irreducible in $\mathbb Z[x]$, so it generates a prime ideal. This shows that the factor ring $\mathbb Z[x]/(x^2+2x+3)$ is an integral domain. However, it's not a field since the ideal $(x^2+2x+3)$ is not maximal: we have $$(x^2+2x+3)\subsetneq(x^2+2x+3,5)\subsetneq\mathbb Z[x].$$

(b) $\mathbb F_5[x]/(x^2+x+1)$ is an integral domain for similar reasons. But $\mathbb F_5[x]$ is a PID (which is not the case with $\mathbb Z[x]$), and in a PID a non-zero prime ideal is maximal; see here.

(c) $\mathbb{R}[x]/(x^4+2x^3 +x^2 +5x+2)$ can't be an integral domain because the polynomial $x^4+2x^3 +x^2 +5x+2$ is reducible over $\mathbb R$ (why?), so the ideal it generates is not prime.