The homogeneous coordinates are very useful to avoid degenerate cases.
A similar construction can be obtained with Cartesian coordinates.

restart;
r:=rand(-12..12):
P,Q,R := 'r()*x+r()*y+r()'$3:
q :=(a,b) -> a*Q*R+b*P*R+P*Q:
plots:-implicitplot( [P, Q, R, q(6,-6)], x=-5..5, y=-5..5, color=[red$3,blue]);

This seems to be about twice faster.

SP:= proc(A::list(set), B::list(set))
local i,j, M:=Array(1..nops(A),1..nops(B),datatype=integer[4]);
for i to nops(A) do for j to nops(B) do
  if A[i] subset B[j] then M[i,j]:=j fi od od;
[seq(subs(0=NULL,[entries(M[i],nolist)]),i=1..nops(A))]
end:

Maybe you should always check the envelope.
In this case,  [y=sin(x+c), cos(x+c)=0] ==> y = +-1.

Note. For the IV problem

dsolve({diff(y(x),x)=sqrt(1-y(x)^2), y(0)=-1});

Maple gives  y(x)=-1, but it has infinitely many solutions in any nbd of 0, e.g.
S:=a -> piecewise(x<a, -1, -cos(x-a));    #  0 <= a,   x<a+Pi
which probably Maple will never (?) find. 

Same result with:

simplify(coeff(series(f,x), x^2));

ex:=sqrt(4*n^2+5*n-7)/n:
sqrt(expand(ex^2)); 

(assuming n>0).

You should be aware that in Maple it could be difficult to obtain the answer in a preferred form.

Maple can solve systems of polynomial inequalities. SemiAlgebraic is more flexible than solve.

sys := r-c>0, 0<=r^2 - 2*r*c -r +2*c, r^2 - 2*r*c -r +2*c  < (r-c)^2, 0<=c, c<=1 , 0<r, r<1, r<2*c:
SolveTools:-SemiAlgebraic([sys], [r], parameters=[c]);

The system has infinitely many solutions:

Sys:=lhs~(sys):
Groebner:-IsZeroDimensional(Sys);  # ==> false

To see some of them:

solve(sys, maxsols=10);

I don't see this as a real (annoying) bug, because sol(parameters)  and sol(parameters=...) is supposed to be used only for display purposes.

Here are two working solutions.

1. Create a file A.mpl and use $include

module A()
$include "B.mpl";
export foo:=proc()
  0;
 end proc;
end module;

and in the worksheet use
read "A.mpl"

2. Remove the export from B.mpl, i.e. it will contain 

boo:=proc()
return 0;
end proc; 

and use in the worksheet:

module A()
export boo,foo;
read "B.mpl";

foo:=proc()
  0;
 end proc;
end module;

You cannot, because they differ by a constant.

Why dou you think that the integral can be expressed in terms of known functions?
Note that even int(sqrt(cos(t)^2+cos(2*t)^2),t)  has a complicated form and int(sqrt(cos(t)^2+cos(3*t)^2),t)  is already too much! 

1. The parantheses are not ignored; they are not displayed, being superfluous in this case, similar to a + (b*c), due to the order of precedence of the Maple operators.

2. Unfortunately Maple does not simplify the set operators; they are used mainly for explicit sets.
If you need to check or simplify expressions in set theory, you can use the Logic package because it works practically in a Boolean algebra.