I have just started learning graph theory not long ago, this is a past year problem and I got the correct answer by chance(True/False questions), wanted to check my understanding on this site.
My concern lies with Q17 and Q19, for which the answers are True and False respectively, as chromatic number = minimal vertex cover = $4$.
For Q17, the general approach to finding the chromatic number of the original graph(which is the minimum number of colours to colour the nodes) will be to first find the lower bound of this minimum number. The lower bound can be found from the largest subgraph of the current graph that is complete, so that we can deduce that the lower bound is actually number of vertices of the subgraph, which is $4$ in this case(the left most complete graph in the diagram).
The upper bound of chromatic number can be found by arranging all degrees of vertices in descending order, enumerating them from 1 to $n$ (where n = number of vertices = $8$ in this case), then find the first enumerated number, $k$ where $k+1 > d_{k}$, which also happens to be $4$, and the 2 bounds will reduce to the chromatic number being 4.
Alternatively, we could try colouring with just 4 colours since there are only 8 vertices.
For Q19, I'm not too sure. I am thinking that I must somehow transform this graph into a bipartite graph, and I assume it's always possible so I simply divide the number of vertices by $2$ to get $4$.
Find a lower bound on vertex cover by finding a matching, I.e. a set of edges no two of which share a vertex. A minimal vertex cover must use at least one end of each of the edges in the matching so a matching of size $k$ witnesses that any vertex cover has size at least $k$. I think I can see a matching of size 4.