Here are the basic notations:
Population mean = $\mu$
Sample mean = $\bar x$
Population variance = $\sigma^2$
E = Estimator
Assumption $E(x_i) = \mu$
$E(x_i)$ is any random variable within the population space.
$$E(\bar x) = E(\frac{1}{n} \sum_{i=1}^n x_i)$$ $$E(\bar x) = \frac{1}{n} \sum_{i=1}^n E(x_i)$$ $$E(\bar x) = \frac{1}{n} \sum_{i=1}^n \mu$$ $$E(\bar x) = \frac{1}{n}. n. \mu$$ $$E(\bar x) = \mu$$
Now,
$$Var(x_i) = E(x_i^2) - (E(x_i)^2)$$ $$\sigma^2 = E(x_i^2) - \mu^2$$ $$ E(x_i^2)= \mu^2 + \sigma^2 $$ $ E(x_i^2)= \mu^2 + \sigma^2 $ Let this be equation 1
Also,
$$var(\bar x) = E(\bar x^2)-(E(\bar x))^2$$ $$E(\bar x^2) = \frac{\sigma^2}{n} + \mu^2$$ $E(\bar x^2) = \frac{\sigma^2}{n} + \mu^2$ Let this be equation 2
Also using the Estimator,
$$E(\bar x^2) = E(\frac{1}{n} \sum_{i=1}^n x_i)^2$$ $$E(\bar x^2) = \frac{1}{n^2} \sum_{i=1}^n E(x_i)^2$$
From equation 1 we know that $ E(x_i^2)= \mu^2 + \sigma^2 $ Hence,
$$E(\bar x^2) = \frac{1}{n^2} \sum_{i=1}^n (\sigma^2+\mu^2)$$ $$E(\bar x^2) = \frac{1}{n^2} \sum_{i=1}^n (\sigma^2+\mu^2)$$ $$E(\bar x^2) = \frac{1}{n^2} .n^2. (\sigma^2+\mu^2)$$ $$E(\bar x^2) = (\sigma^2+\mu^2)$$
How can $E(\bar x^2)$ have two values as shown above and equation 2 Where $E(\bar x^2) = \frac {\sigma^2}{n} + \mu^2$
The line after "Also using the Estimator" is incorrect.
It should be
$$E(\bar x^2) = E\left(\left(\frac{1}{n}\sum_{i=1}^n x_i\right)^2\right) = E\left(\left(\frac{1}{n}\sum_{i=1}^n x_i\right)\cdot\left(\frac{1}{n}\sum_{i=1}^n x_i\right)\right)$$
This isn't equal to
$$E\left(\frac{1}{n}\sum_{i=1}^n x_i^2\right)$$
Particularly because you will get terms like $x_i x_j$, where $i \ne j$ in the first case, which you wouldn't get in the second case.
Everything before this line seems correct though; but everything after this line that uses the equation after "Also using the Estimator" is incorrect.