Do  restart  first and everything will work properly:

restart; 
solve(a = b+5, b); 
solve(a*x^2+b*x+c = 0, x);
                                                  

Addition. Directly in your document I get a very strange result:

I think that you should create a new document that is free of any presetting.

It is true for any real  a  and  b  if  b>0 :

is(arctan(a,b)=arctan(a/b)) assuming real, b>0;
                                               true

Download q1nwtnnonlinearsys_new.mw

Function  procedure solves the problem. Required parameters of the procedure: F is an expression of the variable  x  or a procedure, R is a list of the end points of a segment (the domain of the function), N is the number of points (the ends of the segment  R  are included in this number so  N>=2).  Optional parameters: the equality  output=list  (by default) or   output=plot , Opts  is a list (additional options for plotting). 

Function:=proc(F::{algebraic,procedure}, R::list, N::posint, Eq::`=`:=(output=list), Opts::list:=[])
local f, a, b, h, L;
if F::algebraic then f:=x->F else f:=F fi;
a,b:=op(R); h:=(b-a)/(N-1);
L:=[seq(evalf[3]([a+i*h,eval(f(x),x=a+i*h)]), i=0..N-1)];
print(L);
if Eq=(output=plot) then 
plot([f(x), L], x=a..b, color=[red,blue], op(Opts)) fi;
end proc:

Example of use:

Function(x^2, [-2,2], 9, output=plot, [scaling=constrained, size=[500,500]]);

You can use  plots:-contourplot  command
See also this post  https://www.mapleprimes.com/posts/202222-Contour-Curves-With-Labels

Examples of use:

restart:
f(x,y):=exp(-(x^2+y^2)/3)*cos(2*x*y):
plots:-contourplot(f(x,y), x=-2.5..2.5, y=-2.5..2.5, contours=[seq(-0.4 .. 1, 0.1), 0.95] ,filledregions=true, coloring=["White","DarkViolet"], numpoints=50000); 

ContoursWithLabels(f(x,y), -2.5 .. 2.5, -2.5 .. 2.5, {seq(-0.4 .. 1, 0.1), 0.95},Coloring = [colorstyle = HUE, colorscheme = ["White","DarkViolet"], style = surface]);

  Edit.      

I tried to open the attached file by selecting  Maple Input , but my computer freezes after that.
Try to restore your worksheet as follows:

Menu  File -> Recent Documents -> Restore Backup

plot([BesselJ(0,x), BesselJ(1,x)], x=0..10, color=[red,blue]);

See help on  ?Bessel
 

In order to insert a text into each frame for this animation method, we need to create a new sequence of frames using  plots:-display  command:

restart;
with(plots):
N := 92: 
A := [seq(plot([[.85*sin(t)^3-2+1.25*i/N, .85*(13*cos(t)*(1/15)-(1/3)*cos(2*t)-2*cos(3*t)*(1/15)-(1/15)*cos(4*t)), t = 0 .. Pi*i/N], [-.85*sin(t)^3-2+1.25*i/N, .85*(13*cos(t)*(1/15)-(1/3)*cos(2*t)-2*cos(3*t)*(1/15)-(1/15)*cos(4*t)), t = 0 .. Pi*i/N], [sin(t)^3+2-1.25*i/N, 13*cos(t)*(1/15)-(1/3)*cos(2*t)-2*cos(3*t)*(1/15)-(1/15)*cos(4*t), t = 0 .. Pi*i/N], [-sin(t)^3+2-1.25*i/N, 13*cos(t)*(1/15)-(1/3)*cos(2*t)-2*cos(3*t)*(1/15)-(1/15)*cos(4*t), t = 0 .. Pi*i/N]], color = red, thickness = 5, view = [-2.3 .. 2.4, -1.4 .. 1]), i = 1 .. N)]:
 
A1:=[seq(display(textplot([[-1.9+(i-1)*1.2/N, 0,`Betty`],[1.9-(i-1)*1.2/N, 0,`Alf`]], font=[TIMES,16]), A[i]), i=1..N)]: 
 
display(A1, insequence = true, scaling = constrained, axes=none, title ="Alf & Betty", titlefont=[TIMES,18] );

verify(2*log(x), log(x^2)) assuming x>0;                          
evalb(2*log(x) = log(x^2)) assuming x>0;

                                             true
                                             true

restart:
h1:=z->1-(delta2/2)*(1 + cos(2*(Pi/L1)*(z - d1 - L1))):
h2:=z->1-(delta2/2)*(1 + cos(2*(Pi/L2)*(z - d2 - L2))):
h3:=z->1+(delta2/2):
K1:=((4/h1(z)^4)-(sin(alpha)/F)-h1(z)^2+Nb*h1(z)^4):
K2:=((4/h2(z)^4)-(sin(alpha)/F)-h2(z)^2+Nb*h2(z)^4):
K3:=((4/h3(z)^4)+(cos(alpha)/F)+h3(z)^2+Nb*h3(z)^4):
lambda1:=Int(K1,z=0..0.2):
lambda2:=Int(K2,z=0.2..0.4):
lambda3:=Int(K3,z=0.4..0.6):
lambda:=lambda1+lambda2+lambda3:
 
F:=0.3:
L1:=0.2:
d1:=0.2:
d2:=0.2:
L2:=0.3:
alpha:=Pi/6:
plot( [seq(eval(lambda, Nb=j), j in [0.1,0.2,0.3])], delta2=0.02..0.1);

Try adding this condition before  od;
if  not type(prO[i], float) then break fi;

If I understand the problem correctly, then we must sum those terms of the geometric series  seq((1-p)^(n-1)*p, n=1..infinity)  for which  sin(n)<=1/2  is true. It is unlikely that this can be done symbolically (exactly). Here numerical calculation with high accuracy:

 S:=0: p:=1/3: q:=1-p:
for n from 1 to 100 do
if evalf(sin(n))<=1/2 then S:=S+q^(n-1)*p fi;
od:
evalf(S);  # Result
evalf(2839595/4782969);
evalf(1/3*(2/3)^100/(1-2/3));  # Error estimation

                                                  0.3925416398
                                                  0.5936887736                     
                                             2.459654427*10^(-18)   

You have a rather complex non-linear system with 5 parameters. Maple simply does not know how to solve such systems. If we specify the values of the parameters, then  fsolve  command easily finds a solution:

eq1 := 2*(r/a)^beta*V[0]*r^(4+Omega)-2*alpha*Omega-alpha^2+alpha*beta+2*alpha: 
eq2 := Omega*alpha+Omega*beta-k^2+Omega: 
eq3 := -2*(r/a)^beta*V[0]*r^(4+Omega)+2*Omega^2+alpha*Omega-Omega*beta-2*Omega+Y[0]*(r/a)^(beta-3*Omega)*m^2: 
eq4 := 2*(r/a)^beta*V[0]*r^(4+Omega)*Omega+4*(r/a)^beta*V[0]*r^(4+Omega)+2*k^2*Omega+k^2*alpha-k^2*beta-2*k^2+4*Y[0]*(r/a)^(beta-3*Omega)*m^2: 
Sys := {eq1, eq2, eq3, eq4}; 
Parameters := indets(Sys, name) minus {Omega, alpha, beta, k};
 
fsolve(eval({eq1, eq2, eq3, eq4}, Parameters=~{1, 2, 3, 4, 5}), {Omega, alpha, beta, k});

 Explore command allows you to explore how the roots change when individual parameters change:

Explore(fsolve({eq1, eq2, eq3, eq4}, {Omega, alpha, beta, k}), parameters=[a=0.1..3.,m=0...5., r=0...5.,V[0]=0...5.,Y[0]=0...5.] );

Edit.
 

As an alternative, here is another way to solve the problem. In addition to the spiral of triangles, PadovanSpiral  procedure also returns a spiral from segments (a fat red polyline), and the lengths of these segments are consecutive terms of Padovan series.

restart;
Pad:=rsolve({P(1)=1,P(2)=1,P(3)=1, P(k)=P(k-2)+P(k-3)}, P, 'makeproc'):
seq(Pad(i), i=1..30);   # Example of use   

PadovanSpiral:=proc(n::posint)
local S, T, R, L, M, M1, i, P;
uses plots, plottools;
M:=<cos(2*Pi/3),-sin(2*Pi/3); sin(2*Pi/3),cos(2*Pi/3)>;
M1:=<cos(-Pi/3),-sin(-Pi/3); sin(-Pi/3),cos(-Pi/3)>;
R[1]:=[[0,0],[1/2,-sqrt(3)/2]];
for i from 1 to n do
R[i+1]:=[R[i][2],convert(Pad(i+1)/Pad(i)*(M.convert(R[i][1]-R[i][2],Vector))+convert(R[i][2],Vector),list)];
T[i]:=[R[i][],convert(M1.convert(R[i][2]-R[i][1],Vector)+convert(R[i][1],Vector),list)];
L[i]:=line(R[i][], color=red, thickness=6);
P[i]:=textplot([(`+`(T[i][])/3)[], Pad(i)], font=[times,`if`(i<=3,12,14)]);
od;
display(seq(L[i], i=1..n), seq(polygon(T[i], color=`if`(i::odd,"LightYellow", "LightBlue")), i=1..n), seq(P[i], i=1..n), scaling=constrained, axes=none, size=[800,600]);
end proc:

Example of use:

PadovanSpiral(13);  # The first two turns of Padovan spiral

Padovan.mw

Edit.

It's easy:

int((2*x+1)/(x+1), x=2..6);
subs([8=m, ln(3)=n*ln(3), -ln(7)=k*ln(7)], %);

Addition. After I sent the answer, I noticed the attached file and looked it up. The difference between your two cases is random. For example, for the case  d=0 (your second case), Maple can return a result with one logarithm, and maybe even with three logarithms (see the file below). However, under the conditions  b*d-a*c<>0, b>0, a>=0, c>=0, d>=0 , the result can always be expressed unambiguously through one logarithm:

 int((a*x+d)/(b*x+c), x=2..6) =4*a/b + ((b*d-a*c)/b^2)*ln((c+6*b)/(c+2*b))
 

Download Integral.mw

Edit.