(a)
Solution 1: We will prove this via contradiction: assume that line $IC$ intersects line $MP$ at $Q$ and line $EF$ and $R$, with $R$ and $Q$ not equal to $V$. Let $x = \angle A/2 = \angle IAE$ and $y = \angle C/2 = \angle ICA$. We know that $\overline{MP} \parallel \overline{AC}$ because $MP$ is a midsegment of triangle $ABC$; thus, by alternate interior angles (A.I.A) $\angle MVE = \angle FEA = (180^\circ - 2x) / 2 = 90^\circ - x$, because triangle $AFE$ is isosceles. Also by A.I.A, $\angle MQC = \angle QCA = y$. Furthermore, because $AI$ is an angle bisector of triangle $AFE$, it is also an altitude of the triangle; combining this with $\angle QIA = x + y$ from the Exterior Angle Theorem gives $\angle FRC = 90 - x - y$. Also, $\angle VRQ = \angle FRC = 90 - x - y$ because they are vertical angles. This completes part (a).
Solution 2: First we show that the intersection $V'$of $MP$ with the internal angle bisector of $C$ is the same as the intersection $V''$ of $EF$ with the internal angle bisector of $C.$ Let $D$ denote the intersection of $AB$ with the internal angle bisector of $C,$ and let $a,b,c$ denote the side lengths of $BC, AC, AB.$ By Menelaus on $V'', F, E,$ with respect to $\triangle ADC,$ $$\frac{AF}{FD}\cdot \frac{DV''}{V''C}\cdot \frac{CE}{EA}=-1$$
$$\frac{DV''}{V''C}=-\frac{\frac{bc}{a+b}+\frac{a-b-c}{2}}{\frac{a+b-c}{2}}=\frac{b-a}{b+a}.$$ Similarly, $$\frac{BP}{PD}\cdot \frac{DV'}{V'C}\cdot \frac{CM}{MB}=-1$$ $$\frac{DV'}{V'C}=-\frac{\frac{ac}{a+b}-\frac{c}{2}}{\frac{c}{2}}=\frac{b-a}{b+a}.$$ Since $V'$ and $V''$ divide $CD$ in the same ratio, they must be the same point. Now, since $\frac{b-a}{b+a}>-1,$ $I$ lies on ray $CV.$ $\blacksquare$
Solution 3: By the Iran Lemma, we know $CI, EF, MP$ concur, so Part $\text{A}$ follows easily.
(b)
Solution 1: Using a similar argument to part (a), point U lies on line $BI$. Because $\angle MVC = \angle VCA = \angle MCV$, triangle $VMC$ is isosceles. Similarly, triangle $BMU$ is isosceles, from which we derive that $VM = MC = MB = MU$. Hence, triangle $VUM$ is isosceles.
Note that $X$ lies on both the circumcircle and the perpendicular bisector of segment $BC$. Let $D$ be the midpoint of $UV$; our goal is to prove that points $X$, $D$, and $I$ are collinear, which equates to proving $X$ lies on ray $ID$.
Because $MD$ is also an altitude of triangle $MVU$, and $MD$ and $IA$ are both perpendicular to $EF$, $\overline{MD} \parallel \overline{IA}$. Furthermore, we have $\angle VMD = \angle UMD = x$ because $APMN$ is a parallelogram. (incomplete)
Solution 2: Let $I_A$, $I_B$, and $I_C$ be the excenters of $ABC$. Note that the circumcircle of $ABC$ is the nine-point circle of $I_AI_BI_C$. Since $AX$ is the external angle bisector of $\angle BAC$, $X$ is the midpoint of $I_BI_C$. Now $UV$ and $I_BI_C$ are parallel since both are perpendicular to the internal angle bisector of $\angle BAC$. Since $IX$ bisects $I_BI_C$, it bisects $UV$ as well.
Solution 3: Let $X'$ be the antipode of $X$ with respect to the circumcircle of triangle $ABC$. Then, by the Incenter-Excenter lemma, $X'$ is the center of a circle containing $B$, $I$, and $C$. Because $XX'$ is a diameter, $XB$ and $XC$ are tangent to the aforementioned circle; thus by a well-known symmedian lemma, $XI$ coincides with the $I$-symmedian of triangle $IBC$. From part (a); we know that $BVUC$ is cyclic (we can derive a similar argument for point $U$); thus $XI$ coincides with the median of triangle $VIU$, and we are done.
Solution 4: Let $M_b, M_c$ be the midpoints of arcs $CA, AB$ respectively, and let $T_a$ be the tangency point between the $A$-mixtilinear incircle of $ABC$ and $\Gamma$. It's well-known that $T_a \in XI$, $T_aX$ bisects $M_bM_c$, $AI \perp EF$, and $AI \perp M_bM_c$.
Now, it's easy to see $EF \parallel M_bM_c$, so $IUV$ and $IM_bM_c$ are homothetic at $I$. But $T_aX \equiv IX$ bisects $M_bM_c$, so Part $\text{B}$ follows directly from the homothety.
~ ike.chen