The radius of a sphere is half its diameter, which here is 12, so the radius is 6. The volume of the sphere can be calculated by setting $\displaystyle r = 6$ in the formula:

$\displaystyle V = \frac{4}{3} \pi r ^{3}$

$\displaystyle V = \frac{4}{3} \pi \cdot 6 ^{3} = \frac{4}{3} \pi \cdot 216 = 288 \pi$

75% of this is

$\displaystyle 288 \pi \cdot \frac{75}{100} = 216 \pi$

The formula for volume of a rectangular prism is,

 $\displaystyle V=length*width*height$.  

For this problem, since

$\displaystyle \\height=5 \\length=4 \\width=3$

the solution would be 

$\displaystyle V=5*4*3=60$.

The measure of a right angle is $\displaystyle 90$ degrees.

The measure of an exterior angle of a triangle is the sum of the measures of its remote interior angles, so 

$\displaystyle w = x + y$

$\displaystyle y = w - x= 125 - 81 = 44^{\circ }$

The measure of an exterior angle of a triangle is the sum of the measures of its remote interior angles, so 

$\displaystyle w = x + y = 72 + 43 = 115 ^{\circ }$

In an isosceles triangle, the base angles are the same. Also, the three angles of a triangle add up to $\displaystyle 180^{\circ}$.

So, subtract the vertex angle from $\displaystyle 180^{\circ}$. You get $\displaystyle 116^{\circ}$.

Because there are two base angles you divide $\displaystyle 116^{\circ}$ by $\displaystyle 2$, and you get $\displaystyle 58^{\circ}$.

The measure of an exterior angle of a triangle is the sum of the measures of its remote interior angles, so 

$\displaystyle \angle 3 = \angle 1 + \angle 2$.

We also have the following constraints:

$\displaystyle 45^{\circ } \leq m \angle 1 \leq 50^{\circ }$

$\displaystyle 60^{\circ } \leq m \angle 2 \leq 70 ^{\circ }$

Then, by the addition property of inequalities,

$\displaystyle 45^{\circ } + 60 ^{\circ }\leq m \angle 1 +m \angle 2 \leq 50^{\circ } + 70 ^{\circ }$

$\displaystyle 105 ^{\circ }\leq m \angle 3 \leq 120^{\circ }$

Therefore, the measure of $\displaystyle \angle 3$ must fall in that range. Of the given choices, only $\displaystyle 110^{\circ }$ falls in that range.

The measure of an exterior angle of a triangle is the sum of the measures of its remote interior angles, so 

$\displaystyle \angle 3 = \angle 1 + \angle 2$

or 

$\displaystyle \angle 2 = \angle 3 - \angle 1$

Therefore, the maximum value of $\displaystyle \angle 2$ is the least possible value of $\displaystyle \angle 1$ subtracted from the greatest possible value of $\displaystyle \angle 3$:

$\displaystyle 140^{\circ } - 45^{\circ } = 95 ^{\circ }$

The minimum value of $\displaystyle \angle 2$ is the greatest possible value of $\displaystyle \angle 1$ subtracted from the least possible value of $\displaystyle \angle 3$:

$\displaystyle 130^{\circ } - 50^{\circ } = 80 ^{\circ }$

Therefore, 

$\displaystyle 80^{\circ } \leq \angle 2 \leq 95^{\circ }$

Since all of the choices fall in this range, all are possible measures of $\displaystyle \angle 2$.

A right triangle must have one right angle and two acute angles; this means that no angle of a right triangle can be obtuse. But since $\displaystyle 120^{\circ } > 90^{\circ }$, it is obtuse. This makes it impossible for a right triangle to have a $\displaystyle 120^{\circ }$ angle.

One of the angles of a right triangle is by definition a right, or $\displaystyle 90^{\circ }$, angle, so this is the measure of one of the missing angles. Since the measures of the angles of a triangle total $\displaystyle 180^{\circ }$, if we let the measure of the third angle be $\displaystyle x$, then:

$\displaystyle x + 68 + 90 = 180$

$\displaystyle x + 158 = 180$

$\displaystyle x + 158 - 158= 180 - 158$

$\displaystyle x = 22$

The other two angles measure $\displaystyle 22 ^{\circ }, 90^{\circ }$.