1. No. Because Hamilton graph contains Hamilton loop, and Hamilton loop is still connected after removing any point, that is, there can be no cutting points in Hamilton loop, then we know that a graph with cutting points is not necessarily a Hamilton graph.

2. If these two points are in different connected branches, then the sum of all points in the connected branch is odd, and the sum of points is equal to twice the number of edges, and can only be even, so there is a contradiction. So these two points are on the same connected branch.

3.RR = { & lta，a & gt，& ltb，b & gt，& lta，c & gt，& ltb，d & gt}

s(R)= { & lt； a，a & gt，& ltb，b & gt，& lta，c & gt，& ltc，a & gt，& ltb，d & gt，& ltd，b & gt}

R figure: omitted