A. Let $\angle A_oBB'=\theta.$ Clearly $C_0BB'\cong A_0BB' \implies \angle C_0BB'=\theta$ and all similar statements hold. Let $T$ be touch point of sphere with $ABC;$ we can observe that $$6\theta=(3\pi-(3\pi-(\angle AC_0B+\angle CB_0A+\angle BA_0C)))=\angle ATB+\angle CTB+\angle BTC=2\pi\implies \theta =\pi/3.$$
B. It's suffice to prove that $AC_0,CA_0$ concur. Then lines $AA_0,CC_0$ concur and similarly they concur in pairs with $BB_0,$ but these three lines are noncomplanar and therefore intersect at one point. Let $X=AC_0\cap BB'.$ Note that $$\angle C_0XB=\angle C_0AA'=\pi/3=\angle C_0BX\implies |BX|=|BC_0|=|BA_0|,$$and symmetrically $CA_0$ passes through $X.$
C. By three perpendiculars theorem projections of incenter to $A'B',B'C',C'A'$ coincide with projections of touch point $R$ of sphere with $A'B'C'.$ This touch point is the isogonal conjugate wrt $A'B'C'$ of touch point $S$ of this plane with sphere, which also tangent to extensions of lateral faces. But $\overrightarrow{TS}=\overrightarrow{AA'},$ and moreover angles $ATB,BTC,CTA$ has same measure $\pi-\pi/3=2\pi/3,$ so in fact $R$ is the Isodynamic point of $A'B'C'$ and we are done.