Maple's dsolve does not deal with delay differential equations.  You might look at the following MaplePrimes discussions that have dealt with this topic:

www.mapleprimes.com/forum/differentialequationsdelay

www.mapleprimes.com/forum/delaydifferentialequation

www.mapleprimes.com/forum/delaylogistic

The problem is that p = (1/2)*(n+sqrt(n^2-4*n*sigma^2))/n and n=-((sigma)^(2))/(p*(p-1)) are not equivalent.  If you solve the quadratic
n=-((sigma)^(2))/(p*(p-1)) for p, you get two solutions, one with the +sqrt and one with the -sqrt.  For the values p = .2, phi = .5, mu = 2, sigma = 1
that you used, n = 6.25, and then p = (1/2)*(n-sqrt(n^2-4*n*sigma^2))/n, not (1/2)*(n+sqrt(n^2-4*n*sigma^2))/n.

It seems to be 3 for an empty list + 1 for each position in the list + the sum of the lengths of the members of the list.

If what you have is a rectangular grid of points, use the surfdata command in the plots package. Otherwise, you might look at Paul Weiss's package in www.mapleprimes.com/forum/surfaceplots.

Hmmm, the links there don't seem to be working.  Does anyone know if these files are still available?

> expand(%);

See www.mapleprimes.com/forum/color-several-sets-of-inequalities-on-the-same-graph-0

It seems the RootFinding procedures don't like CylinderD with the variable in the first argument, because Maple doesn't know how to differentiate CylinderD with respect to its first argument.

For a complex root with V=10:

> z:=2.0: W:=(2^(n+3/2)*Pi)/(GAMMA(-n/2)*GAMMA(1/2-n/2)):
  f:=1+((abs(V))^2/W^2)*(CylinderD(n,z)^2)*(CylinderD(n,-z)^2-CylinderD(n,z)^2):
  f10:= eval(f, V=10);

> with(plots): 
  implicitplot([Re,Im](eval(f10, n=s+I*t)), s = 0 .. 15, t = 0 .. 1, gridrefine=2, colour=[red,blue]);

The roots correspond to intersections of the red and blue curves.

> fsolve(f10, n = 3 .. 4+I, complex);

3.412828570+.1737752246*I

It doesn't look to me like there are any real zeros of this function.

> F:= CylinderD(n,2)*CylinderD(n,2*I)+exp(I*n*Pi/2);
  plots[complexplot](F, n=-5 .. 5);

You don't need to do it by "hand". 

> sys:={ diff(Pb(X),X) = .3*(-(1.6+.2*X^2)*sqrt(3)*(1-A(X))^2/(1.6+.4*X^2)-Q)/
  (sqrt(3)*(sqrt(3)*(1-A(X))^2)^2), 
 diff(A(X),X) = (.4*X*(2-(1-A(X))*Pb(X))/A(X)-Pb(X))*(-.86*A(X)^2+.345*A(X)+.515)/
  ((0.8e-1+0.2e-1*X^2)*(2-((2-(1-A(X))*Pb(X))/A(X)-Pb(X))/(2.571-A(X)-A(X)*log(1-A(X))))), 
  Pb(0)= 0, A (0)=.441};
  S:= dsolve(sys,numeric,parameters=[Q]);
  F:= proc(q) S(parameters=[q]); subs(S(1), Pb(X)) end proc;
# thus F(q) is the value of Pb(1) for the solution with Q = q  
   Q0:= fsolve(F);

         Q0 := -.0006338875370

> S(parameters=[Q0]):
  plots[odeplot](S,[[X,Pb(X)],[X,A(X)]], X=0..1, colour=[red,blue], numpoints=1000);

You can get other solutions by specifying an appropriate interval for fsolve.  The graph of F near Q = 0 is quite complicated: it looks like there may be infinitely many solutions.

> plot(F, -.002 .. 0, -5 .. 5, numpoints=1000);

Maple doesn't have "student software" as such.  The "Student Edition" of Maple is the full Maple software at a discounted price.  It contains a Student package which contains a number of things useful to students at various levels.  Yes, it is good.  However, I wouldn't say it's "easy to use".  It's easy to do some things, but to really use it effectively requires a fair amount of learning.

Yes - but it might help if you told us what quantities you want to find the ratio of.  In general, the ratio of A to B is expressed as the quotient A/B

Yes, it is singular at x=0, because the coefficient of the second derivative is 0 there.  This makes numerical methods have trouble with it, and it also means that you have to be very careful about boundary conditions at that point: the conditions for existence and uniqueness are often not what you would expect.

 
Actually, this equation can be solved symbolically:

> S:= dsolve(- diff(u(x),x) - x*diff(u(x),x$2) + u(x)=0);
 
    S := u(x) = _C1*BesselI(0,2*x^(1/2))+_C2*BesselK(0,2*x^(1/2))

The leading behaviour as x -> 0 is as follows:

> series(rhs(S), x = 0, 1);

(_C1 + _C2*(-1/2*ln(x)-gamma)) + O(x)

In order to have u(x) bounded as x -> 0, you need _C2 = 0.  To have u(0) = 1, you then need _C1 = 1.  So your first boundary condition determines a solution:

> u0:= eval(rhs(S), {_C1=1,_C2=0});
 
    u0 := BesselI(0,2*x^(1/2))

However, that will not satisfy your second boundary condition.  In fact, it and all its derivatives are increasing on [0,infty), because its Taylor coefficients are all positive.

Here's one way to use a Slider and Plotter in a Maplet.  It's not really a continuous action, though: to get the plot to change, you must release the mouse button on the slider.

> with(Maplets[Elements]):
  maplet := Maplet([   
   Plotter['PL1']( plot(sin(x), x = 0..10,title=(Phase = 0)) ),
   Slider['SL1'](0..360,filled=true, majorticks=30, minorticks=5, snapticks=false, 
     onchange=Evaluate('PL1'= 'plot(sin(x+'SL1'/180*Pi),x=0..10, title=(Phase='SL1'*degrees))')),
   Button("Done", Shutdown(['SL1']))
   ]):
   Maplets[Display](maplet);

Unfortunately, there doesn't seem to be a way to detect a mouse action in a Plotter.  This has been requested before.

P1 := pointplot([0, 1/4]): P2:=plot(x^2, x = -2 .. 2):
F1 := t -> rotate(translate(P1, L(t)-t, -t^2), -arctan(2*t), 
      [L(t), 0]);
F2 := t -> rotate(translate(P2, L(t)-t, -t^2), -arctan(2*t), 
      [L(t), 0]) end proc;
display([animate(F1, [t], t = -2 .. 2, trace = 10),
         animate(F2,[t], t = -2 .. 2)], scaling = constrained);

It's not at all clear to me what you're trying to do.  You set E₀ to a sequence of two numbers 0, 1.  Is that really what you wanted?  Or did you mean 0.1?