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The vertices $v_2,v_3,v_5$ form a triangle in a graph, so every cut of a graph leaves at least one of their three edges uncut. Hence the cut size is at most four. TakeThe crossing edges are $(1,5),(2,3),(3,5),(4,5)$, so $C=4$. The upper bound is attained and this is a maximum cut.

Choose a partition $V=A\bigsqcup B$ that maximizes the number of edges crossing between its two parts. If some $v\in A$ hadmoving $v$ to $B$ would replace the former crossing edges incident with $v$ by the latter and strictly increase the cut size. This contradicts maximality. The same argument applies to every $v\in B$. Hence every vertex has at least as many neighbours in the opposite part as in its own part, so the maximum cut is an unfriendly partition of a graph.

Every finite graph with $m$ edges has a cut of a graph of size at least $m/2$, and hence a spanning bipartite graph with at least $m/2$ edges. Put each vertex independently on either side with equal probability. Every edge crosses with probability $1/2$, so linearity of expectation gives expected cut size $m/2$.