A new operation $$⊕$$ is defined as $$a⊕b=\frac{2}{a^{2}}+\frac{1}{b}$$:  Which of the equations below is/are correct? 
Overseas Competition->Knowledge Point->Calculation Modules->Operations with New Definition->Operating Directly
method $$1$$:  （$$1$$）$$\frac{2}{2^{2}}+ \frac{1}{4}= \frac{2}{4}+ \frac{1}{4}= \frac{3}{4}$$， $$\frac{2}{4^{2}}+ \frac{1}{2}= \frac{2}{16}+ \frac{1}{2}= \frac{10}{16}= \frac{5}{8}$$. Rejected.  （$$2$$）$$\frac{2}{3^{2}}+ \frac{1}{6}= \frac{2}{9}+ \frac{1}{6}= \frac{4}{18}+ \frac{3}{18}= \frac{7}{18}$$,~ $$\frac{2}{6^{2}}+ \frac{1}{3}= \frac{2}{36}+\frac{1}{3}=\frac{1}{18}+\frac{1}{3}=\frac{1}{18} +\frac{6}{18}=\frac{7}{18}$$.  ($$3$$）$$\frac{2}{4^{2}}+ \frac{1}{8}= \frac{2}{16}+ \frac{1}{8}= \frac{1}{4}$$,  $$\frac{2}{8^{2}}+ \frac{1}{4}= \frac{2}{64}+ \frac{1}{4}= \frac{1}{32}+ \frac{8}{32}= \frac{9}{32} \neq \frac{1}{4}.$$  ($$4$$)$$\frac{2}{1008^{2}}+ \frac{1}{2016}= \frac{1}{252 \times 2016}+ \frac{1}{2016}= \frac{253}{252 \times 2016}.$$  $$\frac{2}{2016^{2}}+ \frac{1}{1008}= \frac{1}{2016 \times 1008}+ \frac{1}{1008}= \frac{2017}{2016 \times 1008}.$$  $$1008⊕2016\neq 2016⊕1008$$.  method $$2$$:  $$4$$ is twice of $$2$$, $$6$$ is twice of $$3$$, $$8$$ is twice of $$4 $$, $$2016$$ is twice of $$1008$$.  Specifically, the question is asking when is $$k⊕2k=2k⊕k$$?  $$\frac{2}{k^{2}}+ \frac{1}{2k}= \frac{2}{\left(2k\right)^{2}}+ \frac{1}{k} \Rightarrow \frac{2}{k^{2}}+ \frac{1}{2k}= \frac{1}{2k^{2}}+ \frac{1}{k}$$  $$\Rightarrow \frac{4}{2k^{2}}+ \frac{1}{2k}= \frac{1}{2k^{2}}+ \frac{2}{2k} \Rightarrow \frac{3}{2k^{2}}= \frac{1}{2k}$$  $$\Rightarrow \frac{3}{k}=1 \Rightarrow k=3$$  Hence, ($$2$$) is the only possible answer. 