could anyone solve this exercice?

Part (a) can be solved without using any property of the circumcenter. This is, let $O$ be an interior point whose cevian triangle $\triangle MNP$ is right at $M$ $(\star).$ Lines $MP$ and $MN$ cut $AC,AB$ at $U,V.$ Since cross ratio $(A,C,N,U)$ is harmonic and $MU \perp MN,$ it follows that $MN,MU$ bisect $\angle AMC$ internally and externally. Similarly, from the harmonic cross ratio $(A,B,P,V)$ and $MV \perp MP,$ we deduce that $MP$ bisect $\angle AMB$ internally. $(\star)$ Locus of point $O$ whose cevian triangle $\triangle MNP$ is M-right, is a quartic $\mathcal{Q},$ but its isogonal conjugate WRT $\triangle ABC$ is a hyperbola $\mathcal{H},$ which passes though the feet of the internal and external angle bisectors of $B,C$ and the isogonal conjugates of the B- and C- vertices of the antimedial triangle.

Construction. Circle with midpoint $O$, see picture. On the y-axis: point $A$. Because $ \angle ABC=60^o$, $ \angle AOC=120^o \rightarrow $ point $C$. $A(0,1),C(-\frac{\sqrt{3}}{2},-\frac{1}{2}), B(\cos a,\sin a)$. Calculating $M=BC \cap AO , N = AB \cap CO , P = AC \cap BO$. Using this coordinates for $MN \bot MP \rightarrow $ equation in $a$, with solution $a : -10^o$ $\Longrightarrow \angle BAC = 70^o$ and $ \angle ACB = 50^o$

Attachments:

(b) Assume that $\angle B=60^{\circ}.$ We already know from part (a) that $MP$ bisects $\angle AMB$ internally. Since $OPBM$ is cyclic, due to $\angle MOP=120^{\circ},$ then it follows that $P$ is the midpoint of the arc $OB$ of $\odot(OMB),$ i.e. $\triangle POB$ is isosceles with apex $P.$ Hence $\angle ABO=\angle POB$ $ \Longrightarrow$ $ 180^{\circ}-2\angle A=90^{\circ}-\angle C=\angle A-30^{\circ} \Longrightarrow \angle A=70^{\circ} \Longrightarrow \angle C=50^{\circ}$

let $\angle BMP=x,\angle PMA=y,\angle AMN=z,\angle NMC=w$ then it's easy to prove $sinxsinz=sinysinw$ if $x>y$,then $w>z$ so $y+z<90$,contradiction. analagously we can reach a contradiction when $x<y$. so $x=y,w=z$.

another solution let $MN$ intersects $AB$ at Q then B,P,A,Q form a harmonic sequence and $\angle PMQ=90$,so it's easy to prove that $MP$ bisects $\angle BMA$.

(a) Simple angle chasing gives $\widehat {AOP}= \pi-2 \beta$, $\widehat {POB}= \pi-2 \alpha$, $\widehat{OAP}=\widehat {OBP}=\frac \pi 2-\gamma$. Using the Sine theorem on triangles $\bigtriangleup APO$ and $\bigtriangleup PBO$ then equating the lenghts of $OP$ give $\frac {AP} {BP}=\frac {\sin (2 \beta)} {\sin (2 \alpha)}$. Moreover, Let $Q:= AB \cap NM$; then the Menelaus theorem in triangle $\bigtriangleup ABC$ with transversal $MN$ gives $\frac {AQ} {BQ}=-\frac {\sin (2 \beta)} {\sin (2 \alpha)}$, so indeed we have $(ABPQ)=-1$. Since $PM \perp MQ$, $PM$ is indeed a bisector of angle $\widehat {AMB}$ and so (by repeating the reasoning) the desired ratio equals $1$. (b) Again, suppose that $\beta=\frac \pi 3$ (other cases are treated in the same way): angle chasing gives $\widehat {AMB}=\frac \pi 2-\beta+\gamma$, $\widehat {AMP}=\frac \pi 2-\gamma$, $\widehat{PMB}=\pi-2\alpha$; using the previously established fact ($MP$ bisector) and $\alpha+\gamma=\frac {2\pi} 3$ allows us to establish easily that $\alpha=\frac {7\pi} {18}$, $\gamma=\frac{5\pi} {18}$.