Because the $X_i$ all have a uniform distribution, all (unordered) variables are assumed independent, and no other order statistic lies between $X_{(1)}$ and $X_{(3)}$, $X_{(2)}$ has a truncated uniform distribution supported on the interval $[X_{(1)}, X_{(3)}]$. Its mean obviously is $(X_{(1)}+X_{(3)})/2$, QED.

If you would like a formal demonstration, note that when the $X_i$ are iid with an absolutely continuous distribution $F$, the conditional density of $X_{(k)}$ (conditional on all the other order statistics) is $dF(x_k)/(F(x_{(k+1)}) - F(x_{(k-1)}))$, which is the truncated distribution. (When $k=1$, $F(x_{0})$ is taken to be $0$; and when $k=n$, $F(x_{n+1})$ is taken to be $1$.) This follows from Joint pdf of functions of order statistics, for instance, together with the definition of conditional densities.

Because the $X_i$ all have a uniform distribution, all (unordered) variables are assumed independent, and no other order statistic lies between $X_{(1)}$ and $X_{(3)}$, $X_{(2)}$ has a truncated uniform distribution supported on the interval $[X_{(1)}, X_{(3)}]$. Its mean obviously is $(X_{(1)}+X_{(3)})/2$, QED.

If you would like a formal demonstration, note that when the $X_i$ are iid with an absolutely continuous distribution $F$, the conditional density of $X_{(k)}$ (conditional on all the other order statistics) is $dF(x_k)/(F(x_{(k+1)}) - F(x_{(k-1)}))$, which is the truncated distribution. (When $k=1$, $F(x_{0})$ is taken to be $0$; and when $k=n$, $F(x_{n+1})$ is taken to be $1$.) This follows from Joint pdf of functions of order statistics, for instance, together with the definition of conditional densities.

Because $X_{(2)}$ has a uniform distribution and is independent of the other variables, and no other order statistic lies between $X_{(1)}$ and $X_{(3)}$, $X_{(2)}$ has a truncated uniform distribution supported on the interval $[X_{(1)}, X_{(3)}]$. Its mean obviously is $(X_{(1)}+X_{(3)})/2$, QED.

If you would like a formal demonstration, note that when the $X_i$ are iid with an absolutely continuous distribution $F$, the conditional density of $X_{(k)}$ is $dF(x_k)/(F(x_{(k+1)}) - F(x_{(k-1)}))$, which is the truncated distribution. (When $k=1$, $F(x_{0})$ is taken to be $0$; and when $k=n$, $F(x_{n+1})$ is taken to be $1$.) This follows from Joint pdf of functions of order statistics, for instance, together with the definition of conditional densities.

Because the $X_i$ all have a uniform distribution, all (unordered) variables are assumed independent, and no other order statistic lies between $X_{(1)}$ and $X_{(3)}$, $X_{(2)}$ has a truncated uniform distribution supported on the interval $[X_{(1)}, X_{(3)}]$. Its mean obviously is $(X_{(1)}+X_{(3)})/2$, QED.

If you would like a formal demonstration, note that when the $X_i$ are iid with an absolutely continuous distribution $F$, the conditional density of $X_{(k)}$ (conditional on all the other order statistics) is $dF(x_k)/(F(x_{(k+1)}) - F(x_{(k-1)}))$, which is the truncated distribution. (When $k=1$, $F(x_{0})$ is taken to be $0$; and when $k=n$, $F(x_{n+1})$ is taken to be $1$.) This follows from Joint pdf of functions of order statistics, for instance, together with the definition of conditional densities.