All the pairs (x, y) that satisfy the inequality
$${2^{\sqrt {{{\sin }^2}x - 2\sin x + 5} }}.{1 \over {{4^{{{\sin }^2}y}}}} \le 1$$
also satisfy the equation

If $$\alpha $$ and $$\beta $$ are the roots of the quadratic equation,
x² + x sin $$\theta $$ - 2 sin $$\theta $$ = 0, $$\theta \in \left( {0,{\pi \over 2}} \right)$$, then
$${{{\alpha ^{12}} + {\beta ^{12}}} \over {\left( {{\alpha ^{ - 12}} + {\beta ^{ - 12}}} \right).{{\left( {\alpha - \beta } \right)}^{24}}}}$$ is equal to :

If a > 0 and z = $${{{{\left( {1 + i} \right)}^2}} \over {a - i}}$$, has magnitude $$\sqrt {{2 \over 5}} $$, then $$\overline z $$ is equal to :

If y = y(x) is the solution of the differential equation
$${{dy} \over {dx}} = \left( {\tan x - y} \right){\sec ^2}x$$, $$x \in \left( { - {\pi \over 2},{\pi \over 2}} \right)$$,
such that y (0) = 0, then $$y\left( { - {\pi \over 4}} \right)$$ is equal to :