Solution:
\( \hat{a} \wedge \hat{b}=\frac{\pi}{4}=\phi \)
\( \hat{a}\cdot\hat{b}=|\hat a||\hat b|\cos\phi \)
\( \hat{a}\cdot\hat{b}=\cos\phi=\frac{1}{\sqrt{2}} \)
\( \cos\theta=\dfrac{(\hat{a}+\hat{b})\cdot(\hat{a}+2\hat{b}+2(\hat{a}\times\hat{b}))}{|\hat{a}+\hat{b}|\,|\hat{a}+2\hat{b}+2(\hat{a}\times\hat{b})|} \)
\( |\hat{a}+\hat{b}|^{2}=(\hat{a}+\hat{b})\cdot(\hat{a}+\hat{b}) \)
\( |\hat{a}+\hat{b}|^{2}=2+2\hat{a}\cdot\hat{b} \)
\( =2+\sqrt{2} \)
\( \hat{a}\times\hat{b}=|\hat a||\hat b|\sin\phi\,\hat{n} \)
\( \hat{a}\times\hat{b}=\frac{\hat{n}}{\sqrt{2}} \) when \( \hat{n} \) is vector \( \perp \) \( \hat{a} \) and \( \hat{b} \)
Let \( \vec{c}=\hat{a}\times\hat{b} \)
We know
\( \vec{c}\cdot\vec{a}=0 \)
\( \vec{c}\cdot\vec{b}=0 \)
\( |\hat{a}+2\hat{b}+2\vec{c}|^{2} \)
\( =1+4+\frac{(4)}{2}+4\hat{a}\cdot\hat{b}+8\hat b+8\hat{b}\cdot\vec{c}+4\vec{c}\cdot\hat{a} \)
\( =7+\frac{4}{\sqrt{2}}=7+2\sqrt{2} \)
Now \( (\hat{a}+\hat{b}).(\hat{a}+2\hat{b}+2\vec{c}) \)
\( =|\hat{a}|^{2}+2\hat{a}\cdot\hat{b}+0+\hat{b}\cdot\hat{a}+2|\hat{b}|^{2}+0 \)
\( =1+\frac{2}{\sqrt{2}}+\frac{1}{\sqrt{2}}+2 \)
\( =3+\frac{3}{\sqrt{2}} \)
\( \cos\theta=\dfrac{3+\dfrac{3}{\sqrt{2}}}{\sqrt{2+\sqrt{2}}\ \sqrt{7+2\sqrt{2}}} \)
\( \cos^{2}\theta=\dfrac{9(\sqrt{2}+1)^{2}}{2(2+\sqrt{2})(7+2\sqrt{2})} \)
\( \cos^{2}\theta=\left(\dfrac{9}{2\sqrt{2}}\right)\dfrac{(\sqrt{2}+1)}{(7+2\sqrt{2})} \)
\( \displaystyle 164 \cos^2 \theta = \dfrac{(82)(9)}{\sqrt{2}} \dfrac{(\sqrt{2} + 1)}{(7 + 2\sqrt{2})} \dfrac{(7.2\sqrt{2})}{(7 - 2\sqrt{2})} \)
\( \displaystyle = \dfrac{(82)}{\sqrt{2}} \dfrac{(9)\left[7\sqrt{2} - 4 + 7 - 2\sqrt{2}\right]}{41}\)
\( =(9\sqrt{2})[5\sqrt{2}+3] \)
\( =90+27\sqrt{2} \)

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