$deg(v)\geq \frac{n}{2}$, then the graph is Hamiltonian.This is Dirac's Theorem
$1)\,deg(v)\geq \frac{n}{3}$.So, this is false.
$deg(v)+deg(w)\geq n$, This is Ore's Theorem
So, $(2)$ is true
The maximum number of edges for a graph to be non-hamiltonian is ${}^{n-1}C_2+1$. For proof refer this
The $(3)$ option says, $E(G)\geq \frac{1}{3}(n-1)(n-2)+2$
For $n=4$,
${}^{4-1}C_2+1=3+1=4$. So, if the number of edges is $\leq4$, then the graph can be non-hamiltonian
$E(G)\geq \frac{1}{3}(4-1)(4-2)+2\geq 4$, this says for $E(G)\geq4$ the graph should be hamiltonian, but for $E(G)=4$ the graph can be non-hamiltonian.
So, $(3)$ should be false.
So, $(B)$ should be the answer,
Correct me if I'm wrong.