For problem (a), 1/3 + 1/12 + 1/12 = 1/2, and 1/2 of a complete revolution is 180º. If a triangle and two dodecahedra meet at a point then their external angles add up to 180º, so the sum of the internal angles is 360º. That tells you that the three polygons lie in a plane. In this case they form part of a tessellation of the plane known as a truncated hexagonal tiling.
For problem (j) there's presumably a misprint in the question. I assume that 1/0 should be 1/10. Then 1/4 + 1/6 + 1/10 > 1/2, so the sum of the internal angles is less than 360º in this case. A square, a hexagon and a decagon can be arranged to meet at a vertex of an Archimedean solid known as a truncated icosidodecahedron.