I don't think we can equate $P(X_1=x_1|X_2=x_2)$ and $E\Big(P(X_1=x_1|X_3)|X_2=x_2\Big)$. To see an example of this, suppose that $(X_1,X_2,X_3)\sim p$ where $p$ is the pmf defined below: $$p(1,1,1)=0.2 \\ p(1,1,2)=0.1 \\ p(1,2,1)=0.01 \\ p(1,2,2)=0.13 \\ p(2,1,1)=0.06 \\ p(2,1,2)=0.11 \\ p(2,2,1)=0.09 \\ p(2,2,2)=0.3 $$ Assume $p(x,y,z)=0$ for all $(x,y,z)\notin \{1,2\}^3$. It's not difficult to verify $$P(X_1=1|X_2=2)=\frac{14}{53}$$ On the other hand, we get with LOTUS that $$\begin{eqnarray*}E\Big(P(X_1=1|X_3)|X_2=2\Big) &=& \sum_{a,b\in \{1,2\}}P(X_1=1|X_3=b)P(X_1=a,X_3=b|X_2=2) \\ &=& \sum_{b\in \{1,2\}}P(X_1=1|X_3=b) \sum_{a\in \{1,2\}}P(X_1=a,X_3=b|X_2=2) \\ &=& \sum_{b\in \{1,2\}}P(X_1=1|X_3=b)P(X_3=b|X_2=2) \\ &=& \frac{7}{12}\cdot \frac{10}{53}+ \frac{23}{64} \cdot \frac{43}{53} \\ &\neq & \frac{14}{53} \end{eqnarray*}$$ If you wish to carry out this computation without the aid of LOTUS (as you started to do) we would need first to establish the conditional pmf of $P(X_1=1|X_3)$ given $X_2=2$. A brief calculator exercise reveals the random variable $P(X_1=1|X_3)$ is supported on the set $\Big\{\frac{7}{12},\frac{23}{64}\Big\}$ and satisfies $$P\Big(P(X_1=1|X_3)=\frac{7}{12}\Big|X_2=2\Big)=P(X_3=1|X_2=2)=\frac{10}{53}$$ $$P\Big(P(X_1=1|X_3)=\frac{23}{64}\Big|X_2=2\Big)=P(X_3=2|X_2=2)=\frac{43}{53}$$ Finally, $$\begin{eqnarray*}E\Big(P(X_1=1|X_3)|X_2=2\Big)&=&\sum_{t\in\big\{\frac{7}{12},\frac{23}{64}\big\}}t P\Big(P(X_1=1|X_3)=t|X_2=2\Big) \\ &=& \frac{7}{12}\cdot \frac{10}{53}+ \frac{23}{64} \cdot \frac{43}{53} \\ &\neq& \frac{14}{53} \end{eqnarray*}$$ As I mentioned in the comments, we may certainly conclude that $$P(X_1=1|X_2=2)=E\Big(P(X_1=1|X_2,X_3)|X_2=2\Big)$$
The mistake is in the line $$P(X_1 > X_2 | X_2 > x_3) = \int_{x_3}^1 P(X_1 > X_2|X_2=x_2)dx_2.$$ You forgot to divide by $P(X_2 > x_3) = 1-x_3$, so the RHS is $P(X_1 > X_2\cap X_2 > x_3)$ rather than $P(X_1 > X_2 | X_2 > x_3)$.
Best Answer
There is no need to work so hard for this question.
Think of it this way. There are only six possible outcomes (except for those outcomes for which at least two of the three random variables are exactly equal, but these occur with probability zero):
$$X_1 < X_2 < X_3 \\ X_1 < X_3 < X_2 \\ X_2 < X_1 < X_3 \\ X_2 < X_3 < X_1 \\ X_3 < X_1 < X_2 \\ X_3 < X_2 < X_1$$ Because each $X_i$ is independent and identically distributed, the probability of any one of these is equal to the other. The underlying distribution makes no difference. Therefore, the answer is $1/6$. Had these not been IID then you'd need to do more work and the answer would depend on how they are distributed, but in this case, the variables are exchangeable.