restart;
plots:-implicitplot3d(min(1/16*(3*x^2+10*x*z-16*y^2+3*z^2), x, z)=0, x = -5 .. 15, y = -15 .. 15, z = -5 .. 15, grid = [100, 100, 100], style = surface, axes=normal);

Comments on solutions:

Problem 2. Note that the ends of these two chords lie on the same circle if and only if the equality  w*x=y*z  is true. We introduce a Cartesian coordinate  X, Y  system in which the origin coincides with the intersection point of the chords, the orientation is normal.

Problem 3. We denote by h the height of the cone, V  is the volume of the filled part, V1  is the volume of the unfilled part. We take advantage of the fact that the ratio of the volumes of similar bodies is equal to the ratio of cubes of the corresponding linear dimensions.

Problem 3. We denote by  [0, y0]  the coordinates of the center of this circle and find  y0  using the discriminant of the obtained quadratic equation.

Download problems_2_3_4.mw

L:= [0,3,0,7]:
k:=0:
for i from 1 to 4 do
if L[i] > 0 then k:=k+1; 
x[k]:= L[i]/2
end if;
end do:
convert(x,list);

                                   [3/2, 7/2]

In Maple 2018.2

sum(sin(Pi*n/2)*sin(n*Pi*(x + 1)/2)*cos(n*Pi*t/2),n=1..infinity, formal);
evalc(convert(%, exp));

                        (-sin((1/2)*Pi*(3*t-x+1))-sin((1/2)*Pi*(t+3*x+1))+sin((1/2)*Pi*(3*t+x-1))+sin((1/2)*Pi*(t-3*x+1))-sin((1/2)*Pi*(t+3*x-1))-2*sin((1/2)*Pi*(t-x-1))+2*sin((1/2)*Pi*(t+x-1))-sin((1/2)*Pi*(3*t-x-1))+2*sin((1/2)*Pi*(t+x+1))-2*sin((1/2)*Pi*(t-x+1))+sin((1/2)*Pi*(3*t+x+1))+sin((1/2)*Pi*(t-3*x-1)))/((2*(exp(I*Pi*(t+x))-1))*(exp(I*Pi*(t-x))-1)*(exp(-I*Pi*(t-x))-1)*(exp(-I*Pi*(t+x))-1))

                                                          0

To calculate the areas integrals are used. If this is not permitted, then use the formula for the area of a circular segment.

Download Plottings_and_calculations.mw

See the corrected

permanouuuuuuuuuuuuuuuuuun_new.mw

Eq := a*x+b*y+c = 0; 
expand((Eq-a*x-c)/b);

L1:=[1,2,3,4,5]:
L2:=[0,5,2,3,7]:
L3:=[7,5,2,3,2]:
convert~([L1,L2,L3], set);
`intersect`(%[]);

                     [{1, 2, 3, 4, 5}, {0, 2, 3, 5, 7}, {2, 3, 5, 7}]
                                             {2, 3, 5}

If necessary, we can take any subsets of this set.

You did not present your example in an editable form, so I am showing the solution with a simple example (terms with derivatives are left on the left side of the equation, the rest are moved to the right):

restart;
Eq:=2*x(t)+1/3*diff(x(t),t)-5=0;
Terms:=[op(lhs(Eq))];
(select,remove)(has,Terms,diff);
add(%[1])=-add(%[2]);

Of course, for use, this code should be written as a procedure, and then you can apply it to each of your differential equations.

restart;
Student:-Precalculus:-CompleteSquare(x^2 + y^2 - 10*x - 75 = 0, [x,y]);
% + 100;

                                    (x-5)^2+y^2-100 = 0
                                     (x-5)^2+y^2 = 100

Unfortunately, the  simplify  command does not always cope with the task. If you want to check the equivalence with one command, then you can use the  is  command:

Download q_new.mw

I wrote down all the code in 1d math (I hate 2d math input) and added the plotting of  x(t)  and  y(t) :

Download Solving_ODE_new.mw

Side surface and bases of a cylinder can be specified parametrically using 2 parameters (as any 2D surface).

Side:=plot3d([cos(t),sin(t),z], t=0..2*Pi, z=0..2, style=surface, color=green):
Base1:=plot3d([R*cos(t),R*sin(t),0], t=0..2*Pi, R=0..1, style=surface, color=green):
Base2:=plot3d([R*cos(t),R*sin(t),2], t=0..2*Pi, R=0..1, style=surface, color=green):
plots:-display(Side,Base1,Base2);

tax := 0.3*profit:
profit := 0.1*totalsalesx:
solve(totalsalesx = 60.1 + tax + profit);

                          69.08045977

If you need an absolutely accurate (symbolic solution) then use fractions:

tax := 3/10*profit: 
profit := 1/10*totalsalesx: 
solve(totalsalesx = 60+1/10 + tax + profit); 
floor(%) %+ frac(%); # Isolation of the whole part

                                    6010/87
                                   69 + 7/87  

Download contours2d_3d.mw