Here is another method based on applying  extrema  command and simplifying the result by  rationalize  and  simplify  commands:

restart;
f:=x->(cos(x)+sqrt(3)*sin(x))/(cos(x)+sin(x)+2);
convert(extrema(f(x), {}, x), list);
Min, Max:=simplify(rationalize(%))[];
evalf([Min,Max]);

Output:

You assigned  o8:=0  and  o8  stands in the denominators.

In Maple 2018 everything is the same.
Do not use  simplify  command, but just differentiate:

with(Physics):
Setup(noncommutativeprefix={P,Q});
diff(Q(t)^2*P(t)*Q(t) + Q(t)*P(t)*Q(t)^2, t);

Output:

Replace  e^(c*v)  by  exp(c*v) . e  is just a symbol in Maple.

This works:

restart:
with(Statistics): 
X := Vector([1, 2, 3, 4, 5, 6], datatype = float):
Y := Vector([2.2, 3, 4.8, 10.2, 24.5, 75.0], datatype = float): 
NonlinearFit(a+b*v+exp(c*v), X, Y, v);

sqrt(sqrt(p^2+1)-1)*sqrt(sqrt(p^2+1)+1);
simplify(combine(%)) assuming positive;

Here is another rather effective method of simplifying expressions containing radicals. Since this expression itself (we denote it by  A ) is obviously positive for any real  p , it is equal to the square root of its square   A=sqrt(A^2) . Therefore, we first square it, then simplify and extract the square root:

A:=sqrt(sqrt(p^2+1)-1)*sqrt(sqrt(p^2+1)+1);
sqrt(simplify(A^2))  assuming p>0;
                         

  Edit.  

If you define a matrix or an array, you should specify ranges for the indexes and just specify non-zero elements. The rest of the elements are automatically considered zeros. 

Your example:

H:=Array((1..3)$4, {(1,1,1,1)= value1, (1,2,2,1) = value2, ...});

See help on  Matrix  and  Array  commands for details.

expr:=``(-y^2+1);
abs(expr);
expand(op(1,%));

restart;
expr := 7*ln(arcsin(x))-(1/2)*ln(x-1)*sin(x)-(1/2)*ln(x+1)+f;
applyrule(ln(t::anything)=ln(abs(t)), expr);

Addition.  Such a calculation may cause a problem due to premature calculation. In this case, you can act as in the example below:

restart;
expr := ''7*ln(arcsin(x))-(1/2)*ln(x-1)*sin(x)-(1/2)*ln(x+1)+f+ln(-2)+ln(-exp(1))'';
applyrule(ln(t::anything)=ln(abs(t)), expr);

Should be  plots:-display  instead of  plot:-display  or  display .  Also should be  plottools:-line  instead of  line .Also  sigma1P1, sigma2P1, tau12P1  are not defined. Submit the complete code if you want a more complete answer.

See:
L[1]:=1:  L[2]:=2:
L; 
                                               L

Unfortunately in Maple there is no command that allows to find the set of values of some function (from one or several variables). But there is a rather simple workaround. Using  maximize  and  minimize  commands, we can solve the problem. It remains to apply the property of a continuous function defined on a connected and compact set: if at two points  x1  and  x2  the function takes certain values  m  and  M  (M>m), then for any intermediate number  m<c<M  there is a point  x0  at which the function takes the value  c .

For the function  (R,z)->R-sqrt(R^2+z^2)  we have:

maximize(R-sqrt(R^2+z^2));
minimize(R-sqrt(R^2+z^2));
                                                         0
                                                     -infinity

Therefore, we can conclude that a given function can take any values <=0 .

Download aero_new.mw

Here is a procedure that does the desired for both constant complex numbers and numbers containing parameters, assuming that these parameters are real numbers. To find the polar angle, the built-in two-argument function arctan(b,a)  was used. This function returns the polar angle for the point  (a,b)  in the range  -Pi .. Pi:

restart;
ExponentialForm:=proc(z)
local Z, a, b;
Z:=evalc(z);
a,b:=Re(Z),Im(Z) assuming real; 
``(simplify(sqrt(a^2+b^2)))*exp(``(simplify(arctan(b,a)))*I) assuming real;
end proc:

Examples of use:

ExponentialForm(1+1*I);
ExponentialForm(1-sqrt(2));
ExponentialForm(exp(I*3*x-x/2)/(1+I));
                         

Here is another version of the same procedure in which the multiplication symbol is explicitly displayed as a bold dot: 

ExponentialForm1:=proc(z)
local Z, a, b, c;
Z:=evalc(z);
a,b:=Re(Z),Im(Z) assuming real; 
subs(`%*`=`.`,simplify(sqrt(a^2+b^2))%*exp(simplify(arctan(b,a))%*I)) assuming real;
end proc:

Example:

ExponentialForm1((1+I)^37);
expand(%);
       

Edit. 

It's just the length of the gradient vector of a scalar function  u(x,y) . Formally, this can be calculated using Student:-VectorCalculus package:

with(Student:-VectorCalculus):
V:=Del(u(x,y));
a:=DotProduct(V,V);
n:=V/sqrt(a);
simplify(DotProduct(n,V));

Addition. For calculation at a specific point it is convenient to use the differentiation operator  D .

Example:
restart;
u:=(x,y)->x^2+y^2;
x0:=1: y0:=2:
sqrt(D[1](u)(x0,y0)^2+D[2](u)(x0,y0)^2);

In addition to the unknown  S , your equation contains 2 parameters  epsilon  and  a . To solve the equation, you must specify numerical values for these parameters.

Example:

Download rootof_new.mw

I can not confirm this result. Everything is working properly on my computer (Maple 2018 on Windows 10):

restart;
g:=x->int(exp(-2*t^2), t=x-1..x)-int(exp(-2*t^2), t=x..x+1);
g(1);
evalf(%);
G:=x->Int(exp(-2*t^2), t=x-1..x)-Int(exp(-2*t^2), t=x..x+1);
evalf(G(1));