*Question 1* We can prove that the statement is true. ($\implies$) Suppose $G$ is bipartite. Then there exists $S \subset V$ such that for all $uv\in E,$ $u\in S \iff v\in V \setminus S$so then $\sum_{v\in S} deg(v) = \sum_{u\in V \setminus S} deg(u)$ By handshaking lemma, $2 |E(G)|= \sum_{v\in V} \deg(v) = 2 \sum_{v\in V} \deg(v) $so $\sum_{v\in S} \deg(v) =|E(G)|.$ $(\Longleftarrow)$ *Question 2* We prove that is statement is true. Suppose $G$ is disconnected. Then $G$ has is the union disjoint subgraphs $G_{1}$ and $G_{2}.$ Suppose $G_{1}$ has $x<n$ vertices then $G_{2}$ has $n-x$ vertices. So $|E(G)|=|E(G_{1})|+|E(G_{2})|\leq {x \choose 2} + {n-x \choose 2}$Now $\begin{aligned}\frac{x(x-1)}2+\frac{(n-x)(n-x-1)}2&=\frac{(n-x)^2-(n-x)+x^2-x}2\\&=\frac{n^2-2nx+x^2-n+x+x^2-x}2\\ &=\frac{n^2-2nx-n+2x^2-x}2\\ &\leq\frac{n^2-3n+2}2={n-1 \choose 2}\end{aligned}$since $\begin{aligned}-2nx-n+2x^2-x&\leq-3n+2\\\iff-2(n-x)x&\leq-2n+x+2\\\iff-2n-2x&\leq-2(n/x)+1+2/x\end{aligned}$If $x=1$ this is clearly true. If $2\leq x<n$ $-2n-2x\leq-2n-2\leq-n+1\leq-2(n/x)+1+2/x.$ *Question 3*