I don't understand why in this problem, they integrated over all the variables (x1, x2, x3) for the top part of the bayes theorem in part c https://www.slader.com/textbook/9780321500465-probability-and-statistics-4th-edition/166/exercises/1/
while in this problem, they integrated only over x1 for the top part of bayes theorem in part c. https://www.slader.com/textbook/9780321500465-probability-and-statistics-4th-edition/166/exercises/3/
The first solution is erroneous. It should indeed be:
$$\mathsf P(X_3\leqslant\tfrac 12\mid X_1{=}\tfrac 14, X_2{=}\tfrac 34)~=~\dfrac{\displaystyle\int_0^{1/2}f(\tfrac 14,\tfrac 34,x_{\small 3})~\mathrm dx_{\small 3}}{\displaystyle\int_0^{1}f(\tfrac 14,\tfrac 34,x_{\small 3})~\mathrm dx_{\small 3}} $$
NB The solution they give accidentally works because $\displaystyle\int_0^1\int_0^1 f(\tfrac 14,\tfrac 34,x_{\small 3})\mathrm d x_{\small 2}\mathrm d x_{\small 1}=f(\tfrac 14,\tfrac 34,x_{\small 3})$ but this is only so because of the domain of the supports. You should not be integrating because the conditions for the variables are exact values, not a range.