It is tempting to think you can exactly mirror the arrangement of weight on the left-hand side, so to have 0 at A, 2kg at B and 1kg elsewhere, however this gives a centre of gravity that is a little way above the centre of the shape, slightly in the direction of A.

Instead, we can use the fact that 15 is the product of 5 and 3. A subset of five weights equally spaced (on a regular pentagon) around the shape will have a centre of gravity at the centre of the shape. Likewise, three weights, equally spaced (on an equilateral triangle) will also be balanced. We are looking to place 16 weights altogether, which suggests we can divide it into two pentagons and two triangles, since 5+5+3+3=16. Since we know O has 2kg, we can place both a pentagon and a triangle on that vertex, but that leaves K, M and N short of a weight. We can then place a pentagon through K and N, and a triangle through M. This completes the given weights on the left-hand side, and also determines the positions of the other eight weights on the right-hand side, such that A, D and G have no weight, B and F have a 1kg weight, and C, E and H have 2kg.

As for the bonus question, with now 17 weights in total, the only way to make this a sum of 3s and 5s is 5+4*3. The only way to do this with five of the positions I to O containing 1kg is to have the pentagon CFILO and the four triangles EJO, AFK, CHM and DIN. This will rearrange the 8 weights on A to H as 10211201, and the positions I and O will have the extra weights on the left-hand side.