From the abelian von Neumann algebra, I see the hyperstonean space as its spectrum（analogy with the C*-algebra). Now I want to see some examples of hyperstonean space.
1) Can you give me a hyperstonean space which is not a stonean space?
2) Does a Cantor set be a hyperstonean space? If not, can you show me some concrete hyperstonean space?
On
Every hyperstonean space is stonean. This corresponds to the fact that every von Neumann algebra is an AW$^*$-algebra.
The Cantor set is stonean, but not hyperstonean. I have read this assertion by experts, but I don't know the proof.
I don't think there are any "explicit" examples of hyperstonean spaces. Hyperstonean spaces are the compact sets such that $C(X)$ is a von Neumann algebra, and that's likely their best characterization.
The second statement in Martin's answer is not quite correct - the Cantor space is totally disconnected/zero dimensional but NOT extremally disconnected/stonean. To see this, let us first be clear about our representation of the Cantor space $C$ as, say, real numbers in the interval $[0,1]$ that have ternary expansions without $1$'s. Take a point $x\in C$ which has infinitely many $0$'s and infinitely many $2$'s in its ternary expansion. Then the points below $x$ in $C$ and those above $x$ in $C$ will both be open and have $x$ in their closures, i.e. the closure of either of these open sets is not open and hence $C$ is not stonean.
The first example of a stonean space that is not hyperstonean was given indirectly by Dixmier. He gave an example of a commutative AW*-algebra $A$ that is not a von Neumann algebra which, as Martin alluded to, means the spectrum/pure states of $A$ are stonean but not hyperstonean (when given the weak* topology induced by $A$). I have never been able to track down the original paper but I have heard that his example goes something like this.
Consider the C*-algebra $B$ of all bounded Borel functions from $\mathbb{R}$ to $\mathbb{C}$. The $f\in B$ with meagre (i.e. countable union of nowhere dense sets) support $\mathbb{R}\setminus f^{-1}\{0\}$ form a closed ideal $I$, so $A=B/I$ is a C*-algebra. Every Borel subset $S$ of $\mathbb{R}$ is, modulo a meagre subset, the same as a unique regular open subset $RO(S)$. So projections in $A$ correspond to regular open subsets of $\mathbb{R}$, which form a complete lattice. Thus $A$ is an AW*-algebra.
However, $A$ is not a von Neumann algebra, and does not even have any non-zero normal positive functionals. To see this note that, as the $RO$ function preserves countable supremums, any normal positive functional on $A$ would induce a (positive countably additive) Borel measure $\mu$ on $\mathbb{R}$ which is zero on all meagre subsets. For each $q\in\mathbb{Q}$ (or any other countable dense subset of $\mathbb{R}$), $\mu(\{q\})=0$ so we can take an open interval $O_q$ containing $q$ such that $\mu(O_q)$ is as small as we like. In particular, if $\mu\neq0$ then we can make the intervals $O_q$ small enough so that $\mu(C)\neq0$, where $C=\mathbb{R}\setminus\bigcup_{q\in\mathbb{Q}}O_q$. But $C$ is meagre (even nowhere dense) by construction, so we have a contradiction.
And lastly, if you want an explicit example of a hyperstonean space then you can simply take any set $X$ with the discrete topology (arbitrary subsets are open), or its Stone-Čech compactification $\beta X$.