Firstly, If L2 is parallel to L1, the gradient of L1 = L2. If we then take the generic equation of any straight line to be: y = mx + c, the m (gradient) of any two parallel lines will be equal!
So even before thinking about what the coordinates (3,13) have to do with this question, we can already say L2 has the equation y = 2x + c.
The coordinates (3,13) have been said to be on the line L2. This means that when y = 13 (on line L2), x = 3. Lets put that into our L2 equation then: 13 = 2(3) + c. This leaves c, which we need to find in order to finish the equation.
13 = 6 + c.
Minus 6 from both sides: 13 - 6 = 6 + c - 6
7 = c
so final equation of L2: y = 2x + 7