1) Assume that $m=a^2+ab+b^2$, because $3|m$, we know that $3|[(a-b)^2+3ab]$, which means that $a\equiv b(mod\ 3)$. Therefore, $x=\frac{b-a}3$, $y=\frac{b+2a}3$ are both positive integers, and $3(x^2+xy+y^2)$$=$$\frac{(b-a)^2}3+\frac{(b-a)(b+2a)}3+\frac{(b+2a)^2}3$$=$$a^2+ab+b^2$. So $x^2+xy+y^2$ is $\frac{m}3$, and $x$ and $y$ are both positive integers. Therefore, $\frac{m}3\in{S}$. 2) Assume that $m=x^2+xy+y^2$, $n=z^2+zt+t^2$$(x,y,z,t\in\mathbb{Z})$, then $mn$$=$$(x^2+xy+y^2)(z^2+zt+t^2)$$=$$(xz+yt+xt)^2+(xz+yt+xt)(yz-xt)+(yz-xt)^2$. Assume that $m_0=xz+yt+xt$, $n_0=yz-xt$, then $mn=m_0^2+m_0n_0+n_0^2(m_0,n_0\in{Z})$. According to the definition of $S$, $mn\in{S}$.