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$\large \dfrac{\color{#D61F06}{2016^2-2008^2}}{\color{#3D99F6}{2013^2-2011^2}}$

Using the algebraic identity, $\color{#20A900}{a^2-b^2}=\color{#624F41}{(a+b)(a-b)}$ ,

= $\large \dfrac{\color{#D61F06}{(2016+2008)(2016-2008)}}{\color{#3D99F6}{(2013+2011)(2013-2011)}}$

= $\large \dfrac{\color{#D61F06}{4024 \times 8 }}{\color{#3D99F6}{4024 \times 2 }}$

= $\large \dfrac{\color{#D61F06}{8}}{\color{#3D99F6}{2}}$

= $\large \color{magenta}{4}$

$\Rightarrow \dfrac{2016^2-2008^2}{2013^2-2011^2}$

Using the algebraic identity, $a^2-b^2=(a+b)(a-b)$ ,

= $\dfrac{(2016+2008)(2016-2008)}{(2013+2011)(2013-2011)}$

= $\dfrac{4024 \times 8 }{4024 \times 2 }$

= $\boxed{4}$

Let $\Large a\normalsize =\color{maroon}{2012}$

And substitute in the given expression

$=\dfrac{(a+4)^2-(a-4)^2}{(a+1)^2-(a-1)^2}=\dfrac{(\color{#D61F06}{a^2}+\color{#3D99F6}{16})+8a-(\color{#D61F06}{a^2}+\color{#3D99F6}{16})+8a}{\color{#D61F06}{(a^2}+\color{#3D99F6}{1})+2a-(\color{#D61F06}{a^2}+\color{#3D99F6}{1})+2a}=\dfrac{16a}{4a}=\color{magenta}{4}$