(−1, 3 ) (i) find polar coordinates (r, θ) of the point, where r > 0 and 0 ≤ θ < 2π.

We know that

The "r" coordinate is the square root of the sum of the squares of the rectangular coordinates.
The "θ" coordinate is the arctangent of the "y" coordinate divided by the "x" coordinate, taking quadrant into consideration.
so

 (-1, 3) 
the point belong to the II quadrant
r=(√((-1)² +3²)> r=√10

θ=arctan(3/1)) = 71.57°
remember that θ belong to the II quadrant
so 
θ=180-71.57> θ=108.43°> 0.6024pi

the answer is
(√10, 0.6024pi)