It is given that,

$$\sum\limits_{k = 1}^{13} {{1 \over {\sin \left( {{\pi \over 4} + {{(k - 1)\pi } \over 6}} \right)\sin \left( {{\pi \over 4} + {{k\pi } \over 6}} \right)}}} $$

Let $$\alpha = {\pi \over 4}$$ and $$\beta = {\pi \over 6}$$. Therefore,

$$\sum\limits_{k = 1}^{13} {{1 \over {\sin (\alpha + k\beta )sin(\alpha + (k - 1)\beta )}}} $$

$$ = {1 \over {\sin \beta }}\sum\limits_{k = 1}^{13} {{{\sin ((\alpha + k\beta ) - (\alpha + (k - 1)\beta ))} \over {\sin (\alpha + k\beta )\sin (\alpha + (k - 1)\beta )}}} $$

$$ = {1 \over {\sin \beta }}\sum\limits_{k = 1}^{13} {(\cot (\alpha + (k - 1)\beta ) - \cot (\alpha + k\beta ))} $$

$$ = {1 \over {\sin \beta }}\{ [\cot (\alpha ) - \cot (\alpha + \beta )] + [\cot (\alpha + \beta ) - \cot (\alpha + 2\beta )] + ...... + [\cot (\alpha + 12\beta ) - \cot (\alpha + 13\beta )]\} $$

$$ = {1 \over {\sin \beta }}(\cot \alpha - \cot (\alpha + 13\beta ))$$

$$ = {1 \over {\sin (\pi /6)}}\left( {\cot {\pi \over 4} - \cot \left( {{\pi \over 4} + {{13\pi } \over 6}} \right)} \right)$$

$$ = 2(1 - 2 + \sqrt 3 ) = 2(\sqrt 3 - 1)$$