Given,

$$\alpha = \tan \left( {{{5\pi } \over {16}}\sin \left( {2{{\cos }^{ - 1}}\left( {{1 \over {\sqrt 5 }}} \right)} \right)} \right)$$

We know, $$2{\cos ^{ - 1}}x = {\cos ^{ - 1}}(2{x^2} - 1)$$

$$\therefore$$ $$2{\cos ^{ - 1}}\left( {{1 \over {\sqrt 5 }}} \right) = {\cos ^1}\left( {2 \times {1 \over 5} - 1} \right) = {\cos ^{ - 1}}\left( { - {3 \over 5}} \right)$$

$$\therefore$$ $$\alpha = \tan \left( {{{5\pi } \over {16}}\sin \left( {{{\cos }^{ - 1}}\left( { - {3 \over 5}} \right)} \right.} \right)$$

$$ = \tan \left( {{{5\pi } \over {16}}\sin \left( {\pi - {{\cos }^{ - 1}}\left( { {3 \over 5}} \right)} \right.} \right)$$

$$ = \tan \left( {{{5\pi } \over {16}}\sin \left( {{{\cos }^{ - 1}}\left( {{3 \over 5}} \right)} \right.} \right)$$

$$ = \tan \left( {{{5\pi } \over {16}}\sin \left( {{{\sin }^{ - 1}}\left( {{4 \over 5}} \right)} \right.} \right)$$

$$ = \tan \left( {{{5\pi } \over {16}} \times {4 \over 5}} \right)$$

$$ = \tan \left( {{\pi \over 4}} \right)$$

$$ = 1$$

$$\therefore$$ $$\alpha = 1$$

Also given,

$$\beta = \cos \left( {{{\sin }^{ - 1}}\left( {{4 \over 5}} \right) + {{\sec }^{ - 1}}\left( {{5 \over 3}} \right)} \right)$$

$$ = \cos \left( {{{\cos }^{ - 1}}\left( {{3 \over 5}} \right) + {{\cos }^{ - 1}}\left( {{3 \over 5}} \right)} \right)$$

$$ = \cos \left( {2{{\cos }^{ - 1}}\left( {{3 \over 5}} \right)} \right)$$

$$ = \cos \left( {{{\cos }^{ - 1}}\left( {2\left. {{{\left( {{3 \over 5}} \right)}^2} - 1} \right)} \right.} \right)$$

$$ = \cos \left( {{{\cos }^{-1}}\left( {{{18} \over {25}} - 1} \right.} \right)$$

$$ = {{18} \over {25}} - 1$$

$$ = {{18 - 25} \over {25}}$$

$$ = - {7 \over {25}}$$

$$\therefore$$ $$\beta = - {7 \over {25}}$$

$$\therefore$$ The quadratic equation with roots $$\alpha$$ and $$\beta$$ is

$${x^2} - (\alpha + \beta )x + \alpha \beta = 0$$

$$ \Rightarrow {x^2} - \left( {1 - {7 \over {25}}} \right)x + 1 \times \left( { - {7 \over {25}}} \right) = 0$$

$$ \Rightarrow 25{x^2} - 18x - 7 = 0$$