What is the meaning and difference of { } and ( ) I know that $(1,2,3) \ne (3,2,1)$ but {1,2,3} = {3,2,1}. But what does it mean actually?
And maybe some explanation on why
1. Ø subset of {0} is true?
2. {x} element of {x} is false?
 A: The round brackets and curly brackets (parentheses and braces) are used in many ways and in many contexts.
But in your case, the round brackets mean the contents are an ordered n-tuple, meaning the order of the contents matters. The curly brackets are an unordered n-tuple, more commonly called a set, where the order of the contents does not matter. There are ways in formal set theory to make the difference more clear, but at this level the description I gave is sufficient. As @quid pointed out, in a set you also do not repeat any of the elements in the list: if you do, the extras are ignored. So, the set $\{1,2,2\}$ is the same as the set $\{1,2\}$, which is of course the same as the set $\{2,1\}$. If I had written round brackets in those lists, they would all have been different from each other.
1) Set $A$ is a subset of set $B$ if every element of $A$ is also in $B$. $Ø$ is the empty set, which has no members. Therefore there is no way you can find a member of $Ø$ that is not a member of $\{0\}$. So, $Ø$ truly is a subset of $\{0\}$. Actually, $Ø$ is a subset of every set. There is no way to disprove it!
2) $x$ is an element of $\{x\}$ but $\{x\}$ is not an element of $\{x\}$. Think of it this way. If you put $x$ in a box, you get a box with $x$ in it. $x$ is in the box, but the box is not in itself. In modern set theory (ZF or NBG), the Axiom of Restriction implies that no set is a member of itself. The famous Russell's paradox is based on the possibility of a set being a member of itself. One way to remove the paradox is to state that no set is a member of itself. This cannot be proved from the other axioms, but it makes sense intuitively.
A: To add to Rory's excellent answer, I want to discuss another point. 
It's reasonable to ask "if the only elementary notions are "set" and "element," and sets don't have order, how do you get ordered pairs?" 
The first part of the answer is to decide what the ordered pair $(a, b)$ should mean, and a rather nice answer is that however you define it in terms of sets, it should have the property that $(a, b) = (c, d) $ implies that $a = c$ and $b = d$ (even if $a$ and $b$ happen to be equal, in which case all four are equal). Having this property means, for instance, that you won't find that $(3, 2) = (2, 3)$, because that would imply (by the $a = c$ rule) that $3 = 2$. 
So here's a definition that has the required property: 
We define  $(a, b)$ to be
$$
\{ \{a\}, \{a, b\} \}.
$$
Let's check that this definition really does have the required property. So we'll assume $(a, b) = (c, d)$ and conclude that $a = c$ and $b = d$. 
Case 1: $a = b$. Then we have
$$
(a, b) = \{ \{a\}, \{a, b\}\} = \{\{a\}\, \{a\}\} \{\{a\}\}.
$$
So 
$$
\{ \{c\}, \{c, d\} \} = \{\{a\}\}.
$$
When two sets are equal, every element of the first is an element of the second, so 
$$
\{c\} \in \{\{a\}\}
$$
hence 
$$
\{c\}  =\{a\}
$$
hence $c = a$, and we've got half our conclusion. The same argument says that 
$$
\{c,d\} \in \{\{a\}\}
$$
hence 
$$
\{c,d\}  =\{a\}
$$
hence 
$$
d \in \{a\}
$$
hence
$$
d = a.
$$
Since we're in the case $a = b$, we conclude that $d = b$. DONE (with case 1). 
In the second case, $a \ne b$. 
We are assuming that $(a, b) = (c, d)$, so 
$$
\{ \{c\}, \{c, d\} \} = \{ \{a\}, \{a, b\} \}.
$$
Hence either case A:
$$
 \{c, d\} =  \{a\}
$$
or case B
$$
 \{c, d\} =   \{a, b\}
$$
must hold. Let's consider case A. Since the two sets are equal, we have
$$
c \in \{a\} \\
d \in \{a\} 
$$
so $c = a$ and $ d = a$, so our original equality becomes
\begin{align}
\{ \{c\}, \{c, d\} \} = \{ \{a\}, \{a, b\} \} \\
\{ \{a\}, \{a, a\} \} = \{ \{a\}, \{a, b\} \}\\
\{ \{a\}, \{a\} \} = \{ \{a\}, \{a, b\} \}\\
\{ \{a\} \} = \{ \{a\}, \{a, b\} \}\\
\end{align}
Hence, since every element of the right-hand set must be an element of the left-hand one, we have
$$
\{a, b\} = \{a\}
$$
and thus $b = a$, which is a contradiction. 
That leaves us in case $B$, in which $a \ne b$ and 
$$
\{a, b\}  =  \{c, d\}.
$$
Because $a \ne b$, we have $\{a\} \ne \{a, b\}$ (you should check this, be a proof very like the ones in the steps above!). Thus $\{a\} \ne \{c, d\}$ (because this latter set is equal to $\{a, b\}$). Hence we must have 
$$
\{a\} = \{c\}
$$
so $a = c$. Our hypothesis for this case now becomes $a \ne b$ and 
$$
\{a, b\} = \{a, d\}.
$$
The element $b$ of the left-hand set must be in the right-hand set. The only two choices are $b = a$ and $b = d$ (or both, if by some chance it turns out that $a = d$...but it won't.)
The first case explicitly contradicts the hypothesis for this case, so the second must hold. We have that $b = d$, and we're done. 
========================================
That's a lot of formal rigamarole just to prove this little property of ordered pairs. The good news is that you never again have to think about it: you've got a new notation, and a definition for it that gives it the property you care about, and you will never look at the proof again (unless, like me, you teach it someday, or write an answer to SE question!). 
