From the given system, we have

$${{k + 1} \over k} = {8 \over {k + 3}} \ne {{4k} \over {3k - 1}}$$

( as System has no solution)

$$ \Rightarrow {k^2} + 4k + 3 = 8k$$

$$ \Rightarrow k = 1,3$$

If $$k = 1$$ then $${8 \over {1 + 3}} \ne {{4.1} \over 2}$$ which is false

And if $$k = 3$$

Then $${8 \over 6} \ne {{4.3} \over {9 - 1}}$$ which is true, therefore $$k=3$$

Hence for only one value of $$k.$$ System has no solution.

$$f\left( x \right) = 2{x^3} + 3x + k$$

$$f'\left( x \right) = 6{x^2} + 3 > 0$$

$$\forall x \in R$$ $$\,\,\,\,\,\,$$ (as $$\,\,\,\,\,\,$$ $${x^2} > 0$$)

$$ \Rightarrow f\left( x \right)$$ is strictly increasing function

$$ \Rightarrow f\left( x \right) = 0$$ has only one real root, so two roots are not possible.

Given equation is $${e^{\sin x}} - {e^{ - \sin x}} - 4 = 0$$

Put $${e^{{\mathop{\rm sinx}\nolimits} \,}} = t$$ in the given equation,

we get $${t^2} - 4t - 1 = 0$$

$$ \Rightarrow t = {{4 \pm \sqrt {16 + 4} } \over 2}$$

$$\,\,\,\,\,\,\,\,\,\,\, = {{4 \pm \sqrt {20} } \over 2}$$

$$\,\,\,\,\,\,\,\,\,\,\, = {{4 \pm 2\sqrt 5 } \over 2}$$

$$\,\,\,\,\,\,\,\,\,\,\, = 2 \pm \sqrt 5 $$

$$ \Rightarrow {e^{\sin x}} = 2 \pm \sqrt 5 $$ $$\,\,\,\,\,$$ (as $$t = {e^{\sin x}}$$)

$$ \Rightarrow {e^{\sin x}} = 2 - \sqrt 5 $$ and

$$\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $${e^{\sin x}} = 2 + \sqrt 5 $$

$$ \Rightarrow {e^{\sin x}} = 2 - \sqrt 5 < 0$$

and $$\,\,\,\,\,\,\sin x = \ln \left( {2 + \sqrt 5 } \right) > 1$$ So, rejected

Hence given equation has no solution.

$$\therefore$$ The equation has no real roots.

$${x^2} - x + 1 = 0$$

$$ \Rightarrow x = {{1 \pm \sqrt {1 - 4} } \over 2}$$

$$x = {{1 \pm \sqrt 3 i} \over 2}$$

$$\alpha = {1 \over 2} + i{{\sqrt 3 } \over 2} = - {\omega ^2}$$

$$\beta = {1 \over 2} - {{i\sqrt 3 } \over 2} = - \omega $$

$${\alpha ^{2009}} + {\beta ^{2009}} = {\left( { - {\omega ^2}} \right)^{2009}} + {\left( { - \omega } \right)^{2009}}$$

$$ = - {\omega ^2} - \omega = 1$$