You ask about three different classes of nonzero ideals in a Dedekind domain $R$:
(a) Prime ideals.
(b) Unfactorable ideals: if $\mathfrak{p} = IJ$ then $I = R$ or $J = R$.
(c) Irreducible ideals: If $\mathfrak{p} = I \cap J$ then $I = \mathfrak{p}$ or $J = \mathfrak{p}$.
Classes (a) and (b) are the same: their equivalence follows easily from the fact that every nonzero ideal in a Dedekind domain factors uniquely as a product of primes.
Class(c) consists precisely of the prime powers $\mathfrak{p}^a$ for $a \in \mathbb{N}$. This follows from the following more general result about intersections of ideals in a Dedekind domain:
Lemma. Let $I = \mathfrak{p}_1^{a_1} \cdots \mathfrak{p}_r^{a_r}$, $J = \mathfrak{p}_1^{b_1} \cdots \mathfrak{p}_r^{b_r}$ be nonzero ideals in a Dedekind domain: here $a_i, b_i$ are non-negative integers. Then $I \cap J = \mathfrak{p}_1^{\max (a_1, b_1)} \cdots \mathfrak{p}_r^{\max (a_r, b_r)}$.
Let me know if you need help in proving this. If you know about localizations, this allows for a very easy proof: you can reduce to the case in which $R$ has exactly one nonzero prime ideal, and in this case the result is trivial.
Finally, one has the following simple result.
Lemma. If $\mathfrak{p}$ is an ideal in a commutative ring $R$, the following are equivalent:
(i) For any ideals $I$, $J$ in $R$, if $\mathfrak{p} \supset IJ$, then $\mathfrak{p} \supset I$ or $\mathfrak{p} \supset J$.
(ii) $\mathfrak{p}$ is prime.
Proof: $\neg$ (i) $\implies$ $\neg$ (ii): The hypotheses give the existence of $x \in I \setminus \mathfrak{p}$ and $y \in J \setminus \mathfrak{p}$ such that $xy \in \mathfrak{p}$. Thus $\mathfrak{p}$ is not prime.
$\neg$ (ii) $\implies$ $\neg$ (i): If $\mathfrak{p}$ is not prime, then there are $x,y \in R \setminus \mathfrak{p}$ with $xy \in \mathfrak{p}$. Take $I = (x)$, $J = (y)$.
I believe this answers all of your questions, but let me know if I missed something.
Firstly, the statement should require that $(a)$ contains a nonzero prime ideal, since all ideals contain $(0)$, which is prime, since you've specified that our ring, call it $R$ is a domain.
First note that $a$ is prime if and only if the ideal $(a)$ is prime. Thus if we have $\newcommand\pp{\mathfrak{p}}\pp\subseteq (a)$, then it suffices to prove that $a\in \pp$, since then $(a)=\pp$, so $(a)$ is prime.
Thus if $af\in \pp$ with $f\not\in\pp$, then $a\in\pp$, and $a$ is prime. Thus if $a$ is not prime, then if $ar\in \pp$, then $r\in\pp$. However, since every element of $\pp$ is of the form $ar$ for some $r\in R$, this tells us that $a\pp=\pp$.
Now since $a\ne 0$ and $a$ is a nonunit, and $\pp\ne R$ is a proper ideal, $a\pp$ contains no irreducibles (if $a\pi \in a\pp$, then since $a,\pi$ are both nonunits, $a\pi$ is not irreducible). Hence since $a\pp=\pp$, $\pp$ contains no irreducibles. However any nonzero prime ideal in a factorization domain contains an irreducible (let $x=\prod_i \pi_i\ne 0\in \pp$, then if none of the irreducible $\pi_i$s was in $\pp$, then since $\pp$ is prime, their product, $x$ wouldn't be in $\pp$ either, contradiction). Thus we have a contradiction, so we've proved our desired claim.
Note actually that we've proven the stronger claim:
Let $R$ be a factorization domain, then if $0\subsetneq \pp \subseteq (a)\subsetneq R$, with $\pp$ prime, then $\pp=(a)$, so $a$ is prime.
That is, we don't need that $a$ is irreducible, just that it is nonzero and not a unit.
Edit: We can do better. An argument at the beginning of chapter 10 in Eisenbud's Commutative Algebra reminded me that we don't need that $R$ is a factorization domain, just that it is a domain. Instead,
once we have $a\pp=\pp$, we apply Nakayama's lemma to conclude that $(1-ax)\pp=0$, for some $ax\in (a)$. However, since $\pp\ne 0$ and $R$ is a domain, we have $1-ax=0$, but we also have that $(a)$ is proper, so $1\not \in (a)$, so $1-ax\ne 0$. This is a contradiction. Thus, in fact, we have
Let $R$ be a domain. If $0\subsetneq \pp \subseteq (a) \subsetneq R$, with $\pp$ prime, then $\pp=(a)$, so $a$ is prime.
Best Answer
Note that $11=(4+\sqrt{5})(4-\sqrt{5})\in\langle 4+\sqrt{5}\rangle$, and so we have the chain of isomorphisms $$\frac{\mathbb{Z}[\sqrt{5}]}{\langle4+\sqrt{5}\rangle}=\frac{\mathbb{Z}[\sqrt{5}]}{\langle4+\sqrt{5},11\rangle}\cong\frac{\mathbb{Z}[t]}{\langle t^2-5,4+t,11\rangle}.$$ Also, $t^2-5=11-(4-t)(4+t)$, whence $\langle t^2-5,4+t,11\rangle=\langle 4+t,11\rangle$, and thus the above ring $\mathbb{Z}[\sqrt{5}]\big/\langle 4+\sqrt{5}\rangle$ is in fact isomorphic to $$\frac{\mathbb{Z}[t]}{\langle 4+t,11\rangle}\cong\frac{(\mathbb{Z}\big/11)[t]}{\langle \bar{4}+t\rangle}.$$ Now, $\mathbb{Z}\big/11$ is a field, so $(\mathbb{Z}/11)[t]$ is a principal ideal domain, and clearly $\bar{4}+t$ is irreducible – thus prime – in $(\mathbb{Z}/11)[t]$. This means that the above ring is a domain, and so $\langle 4+\sqrt{5}\rangle$ is indeed a prime ideal of $\mathbb{Z}[\sqrt{5}]$.