Since $\angle MAN=\angle MEN$ and $\angle MBN=\angle MDN,$ it follows that $\triangle NAB$ and $\triangle NED$ are congruent by ASA $\Longrightarrow$ $\angle ANB=\angle END$ and $NA=NE,$ $NB=ND$ $\Longrightarrow$ isosceles $\triangle NAE$ and $\triangle NDB$ with common apex $N$ are similar. Thus, $\angle DBC=\angle AND=\angle AEF$ implies that the cyclic quadrangles $NAEF$ and $NDBC$ are similar $\Longrightarrow$ $\angle FAN=\angle CDN$ $\Longrightarrow$ $A,F,C,D$ are concyclic. Let the perpendicular to $MN$ at $M$ cut $\gamma_1$ and $\gamma_2$ again at $X,Y.$ According to problem Concyclic, the center of the circumcircle of $AFCD$ is the midpoint of $\overline{XY}.$

I solved the second part in a different manner, although it leads to the same thing. Note that if $O$ is the centre of $\odot(AFCD)\equiv \gamma,$ then $ \angle (OO_1, O_1N)= \angle(AE, AF)$ because $OO_1$ is perpendicular to the radical axis of $\gamma_1$ and $\gamma.$ So, $\angle OO_1N=\angle EAF=\angle END$ and similarly $\angle OO_2N=\angle ANB.$ Since these angles are equal, and also $\angle ANB=\pi-\angle NAB-\angle NBA=\pi-\frac 12(\angle MO_1N+\angle MO_2N)\equiv \alpha;$ therefore the position of $O$ is independent of our selection of $A$ and $D.$ To construct $O$, we can just construct equal angles of $\alpha$ on $NO_1$ and $NO_2,$ one clockwise and the other counterclockwise to meet at the required circumcentre. [asy] import graph; size(15cm, 15cm); real labelscalefactor = 0.55; pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); pen dotstyle = black; real xmin = -16.03, xmax = 27.44, ymin = -10.67, ymax = 16.09; pen qqwuqq = rgb(0,0.39,0); pen ffqqff = rgb(1,0,1); pen dcrutc = rgb(0.86,0.08,0.24); draw(circle((-3.02,0), 5.01), red); draw(circle((6,0), 7.21), red); draw((-7.67,1.85)--(xmax, -0.76*xmax-4), qqwuqq); /* ray */ draw((13.21,0.29)--(xmin, 0.32*xmin-4), gray); /* ray */ draw(circle((2.98,4.01), 10.88), linetype("4 4") + ffqqff); draw((-3.51,4.99)--(13.21,0.29), blue); draw((9.05,6.54)--(-7.67,1.85), blue); draw((0,-4)--(-3.51,4.99), qqwuqq); draw((0,-4)--(9.05,6.54), gray); draw((0,4)--(0,-4), dcrutc); /* dots and labels */ dot((-3.02,0),dotstyle); label([aopsnowrap]"$$O_1$$"[/aopsnowrap], (-3,0.4), NE * labelscalefactor); dot((0,4),dotstyle); label([aopsnowrap]"$$M$$"[/aopsnowrap], (-0.25,4.5), NE * labelscalefactor); dot((6,0),dotstyle); label([aopsnowrap]"$$O_2$$"[/aopsnowrap], (6.25,0.4), NE * labelscalefactor); dot((0,-4),dotstyle); label([aopsnowrap]"$$N$$"[/aopsnowrap], (-0.35,-5.29), NE * labelscalefactor); dot((-7.67,1.85),dotstyle); label([aopsnowrap]"$$A$$"[/aopsnowrap], (-8.67,2), NE * labelscalefactor); dot((13.21,0.29),dotstyle); label([aopsnowrap]"$$D$$"[/aopsnowrap], (13.49,-0.11), NE * labelscalefactor); dot((9.05,6.54),dotstyle); label([aopsnowrap]"$$B$$"[/aopsnowrap], (9,7), NE * labelscalefactor); dot((3.73,-6.84),dotstyle); label([aopsnowrap]"$$C$$"[/aopsnowrap], (3.31,-8.3), NE * labelscalefactor); dot((-3.51,4.99),dotstyle); label([aopsnowrap]"$$E$$"[/aopsnowrap], (-3.22,5.4), NE * labelscalefactor); dot((-3.11,-5.01),dotstyle); label([aopsnowrap]"$$F$$"[/aopsnowrap], (-3.67,-6.13), NE * labelscalefactor); dot((2.98,4.01),dotstyle); label([aopsnowrap]"$$O$$"[/aopsnowrap], (3.31,4.12), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy] $\Box$

It is kind of obvious from angle-chasing that $AFCD$ is cyclic. Let $\Omega_1\in \gamma_1,\ \Omega_2\in \gamma_{2}$ such that $N\Omega_{1}$ is tangent to circle $\gamma_{2}$ and $N\Omega_{2}$ is tangent to circle $\gamma_{1}$. Let $Q$ be the midpoint of $AF$ and $\{R\}=NQ\cap CD,\ \{S\}=NQ\cap \gamma_{2}$. As $\triangle{NAF}\sim \triangle{NDC}$, and $NQ$ is median in $\triangle{NAF}$, it follows easily that $NR$ is symmedian in triangle $\triangle{NCD}$, so the quadrilateral $NCSD$ is harmonic. Intersecting the fascicle $(NC,ND,NS,NN)$ with the line $AF$, we see that we get $N(A,F,Q,NN\cap AF)=-1\Leftrightarrow N\Omega_{1}\parallel AF$, so $AFN\Omega_{1}$ is an isosceles trapezoid, i.e. $AF$ and $N\Omega_{1}$ share the same perpendicular bisector. Analogously, $CD$ and $N\Omega_{2}$ share the same perpendicular bisector, so the center of $AFCD$ is the circumcenter of triangle $\triangle{N\Omega_{1}\Omega_{2}}$, which is, indeed, a fixed point.

Since $DE=AB$, a rotation of $\angle AME$ centered at $N$ will map the triangles $NED, NAB$, so $\triangle DNB\sim\triangle ENA\therefore\angle NEA=\angle DBN\ (\ 1\ )$; from cyclic $AENF, NBDC$, with $(1)$ we get $\angle AFN=\angle NCD$, i.e. $FADC$ is cyclic. Let $O_1,O_2,O$ be the centres of the circles $(AMN), (BMN), (AFCD)$. Easy angle chase shows $AE\parallel CD, AF\parallel BD$; since $NO_1\bot AE, NO_2\bot BD\implies NO_1\parallel OO_2, NO_2\parallel OO_1$ and $O$ is the symmetrical of $N$ w.r.t. midpoint of $O_1O_2$, a fixed point. Best regards, sunken rock

a) Easy by spiral similarity b) We have by Proizvolov's problem ( from Sharygin's book), we have $ OO_1NO_2 $ is a parallelogramm, hence $ O $ is fixed point.