Statement-I    The points $A\left(-1,4\right),B\left(4,3\right),C\left(2,5\right)$  and $D\left(-3,0\right)$  in that order form a rectangle.

Statement-II     The point P(5, − 3) is one of the two points of trisection of the segment joining the
points A(7, − 2) and B(1, − 5)

Statement A:

Given points are $A\left(-1,4\right),B\left(4,3\right),C\left(2,5\right)$ and $D\left(-3,0\right)$

$AB=\sqrt{{\left(4+1\right)}^2+{\left(3-4\right)}^2}=\sqrt{25+1}=\sqrt{26}$

$BC=\sqrt{{\left(2-4\right)}^2+{\left(5-3\right)}^2}=\sqrt{4+4}=\sqrt{8}$

$CD=\sqrt{{\left(-3-2\right)}^2+{\left(0-5\right)}^2}=\sqrt{25+25}=\sqrt{50}$

$AB\ne CD$

So, given points are not vertices of rectangle

Statement B:

$A\left(7,-2\right)$  and $B\left(1,-5\right)$

Let point $P$  divides $AB$  line segment in the ratio $1:2$

$P=\left(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n}\right)$

$=\left(\frac{1\left(1\right)+2\left(7\right)}{3},\frac{1\left(-5\right)+2\left(-2\right)}{3}\right)$

$=\left(\frac{1+14}{3},\frac{-5-4}{3}\right)$

$P=\left(\frac{15}{3},\frac{-9}{3}\right)=\left(5,-3\right)$

$\therefore P\left(5,-3\right)$  one of the trisection point of line segment $\overline{AB}$