a. We can colour the edges as shown. Since the subgraph formed by edges of the same colour connects all 6 vertices but there are only 5 edges, it is a tree and hence has no closed circuit.

b. There are $\binom{n}{2} \ge \frac{n(n-1)}{2}$ edges in total. By the pigeonhole principle, there are $\frac{n(n-1)}{2k}$ edges which are of the same colour. Consider the subgraph formed by these edges. The number of vertices is at most $n$. Since
$$\frac{n(n-1)}{2k} \ge \frac{n(2k+1-1)}{2k} = n,$$
the number of edges of this subgraph is at least the number of vertices. Thus, there must be a closed circuit. This gives a monochromatic closed circuit in $G$.