$$\because$$ $$\overrightarrow a $$ and $$\overrightarrow b $$ be the vectors along the diagonals of a parallelogram having area 2$$\sqrt2$$.

$$\therefore$$ $${1 \over 2}|\overrightarrow a \times \overrightarrow b | = 2\sqrt 2 $$

$$|\overrightarrow a ||\overrightarrow b |\sin \theta = 4\sqrt 2 $$

$$ \Rightarrow |\overrightarrow b |\sin \theta = 4\sqrt 2 $$ ..... (i)

and $$|\overrightarrow a \,.\,\overrightarrow b | = |\overrightarrow a \times \overrightarrow b |$$

$$|\overrightarrow a ||\overrightarrow b |\cos \theta = |\overrightarrow a ||\overrightarrow b |\sin \theta $$

$$ \Rightarrow \tan \theta = 1$$

$$\therefore$$ $$\theta = {\pi \over 4}$$

By (i) $$|\overrightarrow b | = 8$$

Now $$\overrightarrow c = 2\sqrt 2 (\overrightarrow a \times \overrightarrow b ) - 2\overrightarrow b $$

$$ \Rightarrow \overrightarrow c \,.\,\overrightarrow b = - 2|\overrightarrow b {|^2} = - 128$$ ...... (ii)

and $$\overrightarrow c \,.\,\overrightarrow c = 8|\overrightarrow a \times \overrightarrow b {|^2} + 4|\overrightarrow b {|^2}$$

$$ \Rightarrow |\overrightarrow c {|^2} = 8.32 + 4.64$$

$$ \Rightarrow |\overrightarrow c | = 16\sqrt 2 $$ ..... (iii)

From (ii) and (iii)

$$|\overrightarrow c ||\overrightarrow b |\cos \alpha = - 128$$

$$ \Rightarrow \cos \alpha = {{ - 1} \over {\sqrt 2 }}$$

$$\alpha = {{3\pi } \over 4}$$