Angles in pyramid 
We have to find the value of $\cos\theta$.
I tried it alot. But could not able to do it.
Solution given in the book
 A: This works out nicely with vectors.
Place $A, B, C, D$ on the $x$ and $y$ axes. Let $A = (1, 0, 0), B = (0, -1, 0), C = (-1, 0, 0), D = (0, 1, 0)$
$O$ must be on the $z$ axis by symmetry and $O = (0, 0, h)$ The only question is the value of $h$. Since angle $AOB = 45$ degrees, $OA \cdot OB = ||OA|| ||OB|| cos 45$ degrees.
$$(1, 0, -h) \cdot (0, -1, -h) = \sqrt{1^2 + 0 + h^2} \sqrt{0 + 1^2 + h^2} \frac{1}{\sqrt{2}}$$
$$h^2 = (1 + h^2) \frac{1}{\sqrt{2}}$$
$$\sqrt{2} h^2 = 1 + h^2$$
$$(\sqrt{2} - 1) h^2 = 1$$
$$ h = \sqrt{ \frac{1}{\sqrt{2}-1} }$$
Or equivalently, rationalizing.
$$h = \sqrt{\sqrt{2} + 1}$$
We would like the normals to the triangles $OAB $ and $OBC$
$OA = (1, 0, -h), OB = (0, -1, -h), $ and $OC = (-1, 0, -h)$
$OA \cdot OB  = det$
$$\pmatrix{i &j &k\\1 &0&-h\\0&-1&-h} $$
$$= (-h, h, -1)$$ = normal to face $OAB = N_1$
$$OB \cdot OC = det$$
$$\pmatrix{i &j &k\\0 & -1 &-h\\-1& 0 &-h} $$
$$ = (h, h, -1) $$ = normal to face $OBC = N_2$
Then $N_1 \cdot N_2 = (-h, h, -1) \cdot (h, h, -1) = ||N_1||||N_2|| cos \theta$
$$ -h^2 + h^2 + 1 = \sqrt{h^2 + h^2 + 1} \sqrt{h^2 + h^2 + 1}cos \theta$$
$$ cos \theta = \frac{1}{2h^2 + 1}$$
$$cos \theta = \frac{1}{2(\sqrt{2} + 1) + 1} = \frac{1}{2\sqrt{2} + 3}$$
$$ \theta \approx 80.12 degrees$$
As always I do not promise perfect accuracy -- there may be a typo or sign error in there so proofread before you trust it.
But the method is sound and the amount of calculation is reasonable.
Addenda:
Possibly -- please see comment below -- $\theta$ = 180 degrees - 80.12 degrees = 99.88 degrees, 
I plan to build a model and check.
This method is very similar to the text method given ( did not look first :-)  ) but my calculations came out a little differently. As always, double-check first drafts.
A: Let the slant side (= OA = …) be 1 unit.

Find x from the first figure. (x will then be $\sqrt {2 - \sqrt 2}$.)
Next, from the subsequent figures,  find y and z.
Finally apply cosine law to the shaded triangle.
Added: $\theta$, the angle between two planes (OBA and OBC) can only be found by observing the following set of rules:-
(1) Find the line of intersection of the 2 planes (ie. OB);
(2) In plane OBC, find a line perpendicular to OB (Note that many lines qualify but I choose CP.  Then, $\angle OPC = 90^0$);
(3) Repeat step 2 for plane OBA (I use AP);
(4) Test if the lines found in (1), (2) and (3) must be concurrent at the same point. If not, we have to re-do (2) and (3). The choices that I made passed the test and that point is P.
(5) $\theta = \angle CPA$.
A: HINTS:
Shall be brief, hope you fill in details.
Let $AB= 4 b$ then side of inner square on base $ 2 \sqrt 2 b = 2d $ say
Let $OA = h. \, $ Projections on midface and base are $ h \cos \beta, h \sin \beta \,$ where $ 2\beta = AOB$
If $H$ is side of dihedral angle $ 2 \gamma$ between slant faces  
$$ \dfrac{1}{H^2}=\dfrac{1}{(h \cos \beta)^2}+ \dfrac{1}{(h \sin \beta)^2} $$
$$ \sin \gamma= \frac{d}{H} $$
