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? 
 A: 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.
A: Maximal ideals are prime in every ring. Probably you mean the converse, which is true since "divides = contains" in Dedekind domains. So an ideal properly containing a prime ideal would be one properly dividing it, so it can only be $\,(1)$.
A: In any commutative ring with identity, all maximal ideals are prime. An ideal is maximal if and only if the quotient ring is a field; an ideal is prime if and only if the quotient ring is an integral domain. Fields are always integral domains. 
(In fact, in any ring with identity, commutative or not, maximal ideals are prime ideals: if $\mathfrak{A}\mathfrak{B}\subseteq \mathfrak{M}$, with $\mathfrak{M}$ maximal, and $\mathfrak{A}\not\subset\mathfrak{M}$, then $\mathfrak{M}+\mathfrak{A}=R$; hence 
$$\mathfrak{B} = R\mathfrak{B} = (\mathfrak{A}+\mathfrak{M})\mathfrak{B} = \mathfrak{A}\mathfrak{B} + \mathfrak{MB}\subseteq \mathfrak{M},$$
so $\mathfrak{B}\subseteq\mathfrak{M}$, proving that $\mathfrak{M}$ is prime.)
Did you perhaps mean it the other way? 

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

The answer there is "yes" as well, but the reason will depend on which definition of Dedekind domain you have. One definition is that they are domains in which every ideal is finitely generated, every nonzero prime ideal is maximal, and the domain is integrally closed in its field of fractions, which would make the answer true "by definition", but there are other, equivalent, definitions. So if this is what you meant to ask, you'll have to specify what your definition is if you want elucidation. 
