What is difference between undefined & indeterminant form? Then whats about 0^0? I really cant underatand what is exactly difference. By their it can be verified that undefined has no value at all but inderminant form we can able to evaluate its limit by various way(such as L'Hospital's rule) whether it exists.
Indeterminate forms generally refer to combining limits.  If you have two functions $f(x)$ and $g(x)$ such that $\lim_{x \to \infty} f(x)=a, \lim_{x \to \infty} g(x)=b$ you can say $\lim_{x \to \infty} f(x)+g(x)=a+b$.  Other operations can be done similarly.  It is also true that if $\lim_{x \to \infty} f(x)=+\infty, \lim_{x \to \infty} g(x)=b$ you can say $\lim_{x \to \infty} f(x)+g(x)=+\infty$ if you define infinite limits in the usual way.  Indeterminate forms come about when you have something like $\lim_{x \to \infty} f(x)=+\infty, \lim_{x \to \infty} g(x)=-\infty$ where you would like to say $\lim_{x \to \infty} f(x)+g(x)=+\infty-\infty$.  This is indeterminate because you can get any value.  The simple combination is not sufficient and you need to look how fast $f$ and $g$ go off to $\infty$ to come up with the result.  

Undefined means exactly that.  Division by zero is undefined.  $0^0$ is undefined.  This is not in reference to limits at all.
Undefined is something that is not and will never be defined. e.g dividing by zero. Is just an expression that has no sense and no possible value. On the other side, indeterminate is an expression that you can't know its value at simple sight, but it can be anything (even undefined). Some indeterminate forms are $0/0$ or $\infty/\infty$. These kind of expressions are usually solvable using limits, and they can have any value.

In summary, undefined expressions are hopeless, indeterminate ones may have an answer if you work them.

The case $0^0$ is just another example of indeterminate form, and it can be worked in various ways usually taking natural log will bring the exponent down. For example:

$$\lim_{x\to0} x^x=e^{\lim_{x\to0}ln(x^x)}=e^{\lim_{x\to0}xln(x)}=e^{\lim_{x\to0}\frac{ln(x)}{1/x}}$$

So now you can use LH:

$$e^{\lim_{x\to0}\frac{1/x}{-1/x^2}} = e^0=1$$