What is the meaning of $X^{2}$? Can you let me know the meaning of $X^{2}$ where X is a random variable? (Please, give me an example in real life and explain the meaning of $X^{2}$?)

Here's a simple example.  Suppose I give you an unusual six-sided die.  It's not loaded, so any of the six sides is equally likely to be rolled, but the faces are numbered $$1, 1, 1, 2, 2, 3.$$  So, for any given roll of the die, you assign the random variable $X$ to the outcome you observe.  Then you would say $$\begin{align*} \Pr[X = 1] &= \frac{3}{6} = \frac{1}{2}, \\ \Pr[X = 2] &= \frac{2}{6} = \frac{1}{3}, \\ \Pr[X = 3] &= \frac{1}{6}. \end{align*}$$
But what if you were interested not in the number value of the roll, but the square of the value rolled?  You'd first notice that the support of $X$ is the set $\{1, 2, 3\}$; that is to say, the only possible outcomes you can obtain from a single die roll are $\{1, 2, 3\}$.  But the support of $X^2$ is $\{1, 4, 9\}$, because if $X = 1$, then $X^2 = 1$; and if $X = 2$, then $X^2 = 4$; and if $X = 3$, then $X^2 = 9$.  And because we know the value of $X^2$ when we know $X$ itself, it is intuitively obvious that $$\begin{align*} \Pr[X^2 = 1] &= \Pr[X = 1] = \frac{1}{2}, \\ \Pr[X^2 = 4] &= \Pr[X = 2] = \frac{1}{3}, \\ \Pr[X^2 = 9] &= \Pr[X = 3] = \frac{1}{6}. \end{align*}$$  In this situation, we are dealing with a discrete random variable $X$ with finite support, and it so happens that the relationship between $X$ and $X^2$ is one-to-one:  that is to say, $X^2$ is uniquely determined from $X$, but also, $X$ is uniquely determined from $X^2$.  So for example, if you rolled the die and told me that the square of the value you obtained is $X^2 = 9$, I'd tell you that you must have seen $X = 3$.
But now suppose I give you a different die, and although it is also unloaded, the numbers on its faces are $$-1, -1, 1, 1, 2, 2.$$  Then you let $Y$ be the random variable for the value you obtain when rolling this die; you'd state $$\begin{align*} \Pr[Y = -1] &= \frac{1}{3}, \\ \Pr[Y = 1] &= \frac{1}{3}, \\ \Pr[Y = 2] &= \frac{1}{3}. \end{align*}$$  But because $(-1)^2 = 1^2 = 1$, the support of $Y^2$ is only $\{1, 4\}$.  You would say $$\begin{align*} \Pr[Y^2 = 1] &= \Pr[Y = 1] + \Pr[Y = -1] = \frac{2}{3}, \\ \Pr[Y^2 = 4] &= \Pr[Y = 2] = \frac{1}{3}. \end{align*}$$  In this case, if you rolled the die and only told me that $Y^2 = 1$, I would not be able to tell whether the value you rolled was $1$ or $-1$, because both of these outcomes would give the same squared value.
When we deal with continuous random variables, or other transformations of random variables besides squaring, the basic idea is much the same as in the discrete-valued example above.  For a general random variable $X$ following some parametric probability distribution, and some function $f$ that is well-defined on the support of $X$, the expression $f(X)$ is also a random variable, whose support is the image of the support of $X$ under $f$.  For instance, if the support is $$X \in \{ x \in \mathbb R : -1 \le x \le 1 \},$$ and $f(x) = 3x + 5$, then $$f(X) \in \{y \in \mathbb R : 2 \le y \le 8\}.$$
