Moment of Inertia

Removing a $90^\circ$ arc leaves $\frac{3}{4}$ of the ring: $M_{remain} = \frac{3}{4}M$ All of the remaining mass is at distance R from the axis: $I = M_{remain}R^2 = \frac{3}{4}MR^2 \Rightarrow K = \frac{3}{4}$

About a diameter of a thin spherical shell: $I = \frac{2}{3}mr^2$, so $I_1 = \frac{2}{3}mr^2$.
About a tangent (parallel axis theorem): $I = \frac{2}{3}mr^2 + mr^2 = \frac{5}{3}mr^2$, so $I_2 = I_3 = \frac{5}{3}mr^2$.
$I_{XX'} = \frac{2}{3}mr^2 + \frac{5}{3}mr^2 + \frac{5}{3}mr^2 = 4mr^2$

Work required $= \frac{1}{2}I\omega_0^2$, which is minimum when I is minimum, i.e. when the axis passes through the centre of mass.
$m_1x = m_2(L - x)$
$x = \frac{m_2L}{m_1 + m_2}$

Parallel axis theorem: $I = I_{cm} + Md^2$, where d is the distance of the axis from the centre.
I is maximum for the point farthest from the centre.
B lies on the rim (d = R), while A, C and D are closer to the centre, so the moment of inertia is maximum about the axis through B.

Moment of inertia of a disc about its axis: $I = \frac{1}{2}(\text{mass})(\text{radius})^2$
Complete disc: $I_1 = \frac{1}{2}(9M)R^2 = \frac{9MR^2}{2}$
Removed disc (same axis, since it is concentric): $I_2 = \frac{1}{2}M\left(\frac{R}{3}\right)^2 = \frac{MR^2}{18}$
Remaining disc: $I = I_1 - I_2 = \frac{9MR^2}{2} - \frac{MR^2}{18} = \frac{81MR^2 - MR^2}{18}$
$I = \frac{80MR^2}{18} = \frac{40}{9}MR^2$

Moment of inertia of one rod about an axis through its own centre and perpendicular to its length: $I_{cm} = \frac{Ml^2}{12}$
The centre of the square is at a perpendicular distance $\frac{l}{2}$ from the centre of each rod.
By the parallel axis theorem, for one rod about the axis through the centre of the square: $I_1 = \frac{Ml^2}{12} + M\left(\frac{l}{2}\right)^2 = \frac{Ml^2}{12} + \frac{Ml^2}{4} = \frac{Ml^2}{3}$
For the four rods: $I = 4\times\frac{Ml^2}{3}$
$I = \frac{4}{3}Ml^2$

After bending, each half is a thin rod of mass $\dfrac{M}{2}$ and length $\dfrac{L}{2}$.
The axis passes through the bending point, which is one end of each half, and is perpendicular to both halves.
Moment of inertia of a thin rod of mass $m$ and length $l$ about a perpendicular axis through its end $= \dfrac{ml^2}{3}$
For one half: $I_1 = \dfrac{(M/2)(L/2)^2}{3} = \dfrac{ML^2}{24}$
Moment of inertia of the system: $I = \dfrac{(M/2)(L/2)^2}{3} + \dfrac{(M/2)(L/2)^2}{3}$
$I = \dfrac{ML^2}{24} + \dfrac{ML^2}{24}$
$I = \dfrac{ML^2}{12}$
The angle between the halves does not matter, because every element stays at the same distance from the axis.

Moment of inertia: $I = \sum mr^2$. Mass placed farther from the axis contributes much more, since it is multiplied by $r^2$.
The density of iron (about 7.8 g/cm$^3$) is more than that of aluminium (about 2.7 g/cm$^3$).
So to make I maximum, the heavier material (iron) should be placed far from the axis, i.e. on the outer ring, and the lighter aluminium in the interior.
Putting iron sheets on the two flat faces does not move mass farther from the axis.
So the moment of inertia is maximum with aluminium at the interior and iron surrounding it.