Let $O,H$ be the circumcenter and ortocenter of $\triangle ABC.$ If $\angle AMB=2\angle C,$ then $M \in \odot(OAB).$ Since $M$ lies between $O$ and $H,$ then $H$ is outside $\odot(OAB)$ $\Longrightarrow$ $\angle AHB<\angle AOB$ $\Longrightarrow$ $180^{\circ}-\angle C<2\angle C$ $\Longrightarrow$ $\angle C>60^{\circ}.$ Barycentric equation of the circle passing through $A,B$ and $O(a^2S_A:b^2S_B:c^2S_C)$ is $\odot(OAB) \equiv a^2yz+b^2zx+c^2xy-\frac{a^2b^2}{2S_C}z(x+y+z)=0$ Thus, coordinates $(1:1:1)$ of $M$ satisfy the latter equation $2S_C(a^2+b^2+c^2)-3a^2b^2=(a^2+b^2-c^2)(a^2+b^2+c^2)-3a^2b^2=0$ $a^4+a^2b^2+a^2c^2+a^2b^2+b^4+b^2c^2-a^2c^2-b^2c^2-c^4-3a^2b^2=0$ $\Longrightarrow \ c^4=a^4+b^4-a^2b^2$

Medians $AM, BM$ cut circumcircle $(O)$ of $\triangle ABC$ again at $X, Y.$ $M \in \odot(AOB)$ $\Longrightarrow$ $MO$ bisects $\angle BMA$ externally $\Longrightarrow$ $ABXY$ is isosceles trapezoid with diagonal intersection $M$ and equal diagonals $AX = BY$ $\Longrightarrow$ power of $M$ to $(O)$ is $\frac{_4}{^9} m_am_b = AM \cdot BM = R^2 - OM^2 = \frac{_1}{^9}(a^2 + b^2 + c^2)$. Squaring: $(2b^2 + 2c^2 - a^2)(2c^2 + 2a^2 - b^2) = (a^2 + b^2 + c^2)^2$ $\Longleftrightarrow$ $4b^2c^2 + 4a^2b^2 - 2b^4 + 4c^4 + 4c^2a^2 - 2b^2c^2 - 2c^2a^2 - 2a^4 + a^2b^2 =$ $a^4 + b^4 + c^4 + 2b^2c^2 + 2c^2a^2 + 2a^2b^2$ $\Longleftrightarrow$ $3c^4 = 3a^4 + 3b^4 - 3a^2b^2$

(a) Let $A=(0, 0)$, $B=(1, 0)$, $C=(c, d)$; then $M=\left ( \frac {c+1} 3, \frac d 3 \right )$, $O=\left ( \frac 1 2, \frac {c^2+d^2-c} {2d} \right )$. The condition implies that $A$, $B$, $M$, $O$ are concyclic; the circle passing through $A$, $B$, $M$ has equation $\left (x-\ell \right )^2+\left(y-\frac 1 2 \right )^2=\ell ^2+\frac 1 4$ where $\ell=\frac {c^2+2c+d^2-3d+1} {6(c+1)}$; plugging in the coordinates of $O$ we get a multinomial which has $c^4-2c^3+2c^2 d^2+5c^2-2cd^2-4c+d^4+d^2-1$ as a factor, and this is equivalent to the thesis (that is, $1=(c^2+d^2)^2+((c-1)^2+d^2)^2-(c^2+d^2)((c-1)^2+d^2)$ once the other possibilities are excluded. (b) It suffices to note that, from the previous point, the triangle with sides $AB^2$, $BC^2$, $CA^2$ has an angle of $60 ^\circ$.

Let $L$ the symmedian point of $ABC$, and $X \equiv AL \cap BC, Y \equiv BL \cap AC \Rightarrow XLYC$ is cyclical (angle-chasing) By Steiner's Theorem we obtain $BX=\frac{ac^2}{b^2+c^2}$ and $AY=\frac{bc^2}{a^2+c^2}$, we use Lemma $\mathcal{P} _B + \mathcal{P}_A=c^2$ (power with respect to $\odot (CYLX)$) $\Rightarrow \frac{a^2c^2}{b^2+c^2}+\frac{b^2c^2}{a^2+c^2}=c^2 \Rightarrow c^4=a^4+b^4-a^2b^2$ $a^4+b^4-a^2b^2 \ge (a^2+b^2-ab)^2 =a^4+b^4+3a^2b^2-2ab(a^2+b^2)$ $\Longleftrightarrow ab(a^2+b^2) \ge 2a^2b^2$ it's true, then $c^2 \ge a^2+b^2-ab \Rightarrow \angle C \ge 60^{\circ}$

Got it. Part 1 done by wolfram alpha. Hats off to those who solved it manually.

Gravity center is the centroid of $\triangle ABC$. Idk why they chose to frame the question in terms of that instead of using the word centroid though , quite unusual.