a) Note that $C'D'$ is the perpendicular bisector of $AB$, so similarly $A''B''$ is the perpendicular bisector of $C'D'$, so $AB\parallel A''B''$. Similarly $BC\parallel B''C'', CD\parallel C''D'',DA\parallel D''A'',AC\parallel A''C'', BD\parallel B''D''$, thus $ABCD\sim A''B''C''D''$. b)I give up...

Let $\alpha, \beta, \gamma, \delta$ be angles of $ABCD$ at $A, B, C, D$ and $a, b, c, d$ circumradii of $\triangle BCD, \triangle CDA, \triangle DAB, \triangle ABC.$ $[ABCD] = \frac{_1}{^2} AC \cdot BD \cdot \sin \widehat{(AC,BD)}$ $B'D' = \left|b \cos \beta + d \cos \delta \right |= \left| \frac{AC}{2\sin \beta} \cos \beta + \frac{AC}{2\sin \delta } \cos \delta \right | = AC \cdot \frac{\left |\sin (\beta + \delta ) \right|}{2\sin \beta \sin \delta }$ $A'C' = \left|a \cos \alpha + c \cos \gamma \right |= \left| \frac{BD}{2 \sin \alpha} \cos \alpha+ \frac{BD}{2 \sin \gamma} \cos \gamma \right |=$ $ BD \cdot \frac{ \left| \sin (\alpha+ \gamma) \right |}{2 \sin \alpha\sin \gamma}$ $\sin \widehat{(A'C',B'D')} = \sin \widehat{(AC,BD)}$ $[T(ABCD)] =[A'B'C'D'] = \frac{_1}{^2} A'C' \cdot B'D' \cdot \sin \widehat{(A'C',B'D')} = $ $=\frac{_1}{^2} AC \cdot BD \cdot \sin \widehat{(AC,BD)} \cdot \frac{\left| \sin (\alpha+ \gamma) \sin (\beta + \delta)\right |}{4 \sin \alpha \sin \beta \sin \gamma \sin \delta}=[ABCD] \cdot k$ Let $\alpha', \beta', \gamma', \delta'$ be angles of $A'B'C'D'$ at $A', B', C', D'$ $\Longrightarrow$ $\sin \alpha' = \sin \gamma, \sin \beta' = \sin \delta, \sin \gamma'= \sin \alpha, \sin \delta'= \sin \beta$ and $\left| \sin (\alpha' + \gamma') \right| = \left|\sin((\beta' + \delta') \right| = \left |\sin((\alpha + \gamma) \right| = \left |\sin((\beta + \delta) \right|$ $\Longrightarrow$ $[T(T(ABCD))] = [T(A'B'C'D')] = [A''B''C''D''] =[ABCD] \cdot k^2$ $k$ is the similarity coefficient of $A''B''C''D'' \sim ABCD.$