$$\begin{align} \{1,2,3,4,5\}&:=\text{population}\\ \mu&:= \underbrace{\left({ 1+2+3+4+5\over 5 } \right)}_{\text{population mean} } =3\\ \sigma^2&:= \underbrace{\left({1 \over 5} \left(\sum_{i=1}^{5}i^2-5\mu^2 \right) \right)}_{\text{population variance} } =2 \end{align}$$
Sampling without replacement is done for the population with taking 2 elements from it.
$$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|} \hline \text{samples} & (1,2)&(1,3) &(1,4) &(1,5) &(2,3) &(2,4) &(2,5) &(3,4) &(3,5) &(4,5)\\ \hline \bar x & 1.5&2 &2.5 & 3& 2.5& 3& 3.5& 3.5 &4 &4.5 \\ \hline \end{array}$$
The sample-distribution is as following.
$$ \begin{array}{|c|c|c|c|c|c|c|c|}\hline \bar x& 1.5 & 2 & 2.5 & 3 & 3.5 & 4 & 4.5\\\hline P(\overline X=\bar x)& {1 \over 10 } & {1 \over 10 } & {2 \over 10 } & {2 \over 10 } & {2 \over 10 } & {1 \over 10 } & {1 \over 10 } \\ \hline \end{array} $$
$$\begin{align} E(\overline X)=3 ~~ \color{red}{\left(\text{From the symmetry of the distribution} \right)} \end{align}$$
Currently I can't get the meaning of that red-marked statement.
What actually does it be meant?
You can make a graph of the pmf.
The expected value can be considered as follows: You have a graph of the pdf (continous) or pmf (discrete). The x-axis is a kind of seesaw and the bars can be seen as weights. Now you evaluate where you have to put the fulcrum so that the seesaw is in balance. Since the distribution is symmetric the fulcrum must be at the (middle) of the bar in the center. This is at $x=3$. The same considerations can be made for the normal distribution, for instance. For symmetric distributions the center is always the expected value.