$$\widehat a\,.\,\widehat b = {1 \over {\sqrt 2 }}$$ and $$|\widehat a \times \widehat b| = {1 \over {\sqrt 2 }}$$

$${{\left( {\widehat a + \widehat b} \right)\,.\,\left( {\widehat a + 2\widehat b + 2\left( {\widehat a \times \widehat b} \right)} \right)} \over {\left| {\widehat a + \widehat b} \right|\left| {\widehat a + 2\widehat b + 2\left( {\widehat a \times \widehat b} \right)} \right|}} = \cos \theta $$

$$ \Rightarrow \cos \theta = {{1 + 3\widehat a\widehat b + 2} \over {|\widehat a + \widehat b||\widehat a + 2\widehat b + 2(\widehat a \times \widehat b)|}}$$

$$|\widehat a + \widehat b{|^2} = 2 + \sqrt 2 $$

$$|\widehat a + 2\widehat b + 2(\widehat a \times \widehat b){|^2} = 1 + 4 + 4|\widehat a \times \widehat b{|^2} + 4\widehat a\widehat b$$

$$ = 5 + 4\,.\,{1 \over 2} + {4 \over {\sqrt 2 }} = 7 + 2\sqrt 2 $$

So, $${\cos ^2}\theta = {{{{\left( {3 + {3 \over {\sqrt 2 }}} \right)}^2}} \over {(2 + \sqrt 2 )(7 + 2\sqrt 2 )}} = {{9\sqrt 2 (5\sqrt 2 + 3)} \over {164}}$$

$$ \Rightarrow 164{\cos ^2}\theta = 90 + 27\sqrt 2 $$