Hi

You did the wrong way

a) with point (2,-3) and parallel to l2: (x,y)=(1,3)+t(5,2)

(5,2) is a direction vector of l2 and of l1 since l1 is parallel to l2

A Cartesian equation of l1 is therefore 2x-5y+c=0

Substituting (2,-3) gives 2x-5y-19=0

You can check very easily

l1 equation is (x,y)=(2,-3)+t(5,2)

Take t=-1 => (x,y)=(-3,-5) and 2(-3)-5(-5)-19=0

Remember that a line whose equation is ax+by+c=0 is parallel to the vector (-b,a) and perpendicular to the vector (a,b)

b) I let you correct your answer