Is it true that, in a Dedekind domain, all maximal ideals are prime? Is it true that, in a Dedekind domain, all maximal ideals are prime? 

It is true in every ring that maximal ideals are prime (because the quotient is a field, hence also an integral domain).
What is true in Dedekind domains however (by definition) is that every nonzero prime ideal is maximal. See also the Wikipedia article. http://en.wikipedia.org/wiki/Dedekind_domain
Edit:
To add a little more substance to this answer, here are two examples of Dedekind domains. First, the obvious one: $\mathbb{Z}$. Indeed, for every prime ideal $(p)$, we have $\mathbb{Z}/(p)$ is a finite integral domain, hence a field. So every non-zero prime ideal is maximal. It is easy to see that $\mathbb{Z}$ is Noetherian (in fact, of dimension 1), and integrally closed (as is any PID).
Second example: $\mathbb{Z}[i]\cong \mathbb{Z}[x]/(x^2+1)$. By Hilbert's basis theorem, this is Noetherian (overkill-argument!), it is integrally closed (since it is the number field of $\mathbb{Q}(i)$.
Note that the algebraic closure of $\mathbb{Z}$, namely $\bar{\mathbb{Z}}$ is not a Dedekind domain. Unique factorization fails miserably: $a=\sqrt{a} \sqrt{a}=\sqrt{\sqrt{a}}\sqrt{\sqrt{a}} \sqrt{\sqrt{a}}\sqrt{\sqrt{a}}$ and so on. Another way to see it is to consider the ideals $(a)^{\frac{1}{i}}$ for $i=1,2,3,\ldots$. This is a non-terminating increasing chain of ideals, hence $\bar{\mathbb{Z}}$ is not Noetherian.
