For the solution, I used formulas for linear shooting method from the wiki

Addition: the comparison in table form:

C1:=<approx, seq(Y[i], i=1..n+1)>:
C2:=<exact, seq(evalf(eval(exact_sol)), x=X)>:
C3:=C1-C2:
<C1|C2|C3>;
                          

Download LSM1.mw

Edit.

This is probably a bug. Here is a workaround:

restart;
int(exp(-t)/(1-t), t = epsilon .. infinity, CauchyPrincipalValue = true);
limit(%, epsilon=0);
evalf(%);
                         

Here is the solution of Problem 1. A comparison of the graphs shows that the exact solution specified in the task is incorrect.

Download shooting.mw

We can easily prove by hand even an more general result, that for any number of roots in the first term (starting from 2) the limit is 1/2.
We denote by  a[n]  the expression with  n  roots. For example  a[3]=sqrt(x+sqrt(x+sqrt(x)))

It's obvious that  for any  n>=2  we have   
a[n]=sqrt(x+a[n-1]) ,  limit(a[n]/sqrt(x), x=infinity)=1,  limit(a[n]/x, x=infinity)=0

Below we prove that for any  n>=2  the result holds  limit(a[n]-sqrt(x), x=infinity) = 1/2 

We denote  A=sqrt(x+a[n-1]) - sqrt(x) ,  B=sqrt(x+a[n-1]) + sqrt(x)

For the proof, it is sufficient to make only 3 steps: multiply and divide  A  by  B , simplify in the numerator using the difference formula for the squares, and then divide the numerator and the denominator by  sqrt(x)  (below is an illustration of these steps in Maple) :

restart;
A:=sqrt(x+a[n-1])-sqrt(x);
B:=sqrt(x+a[n-1])+sqrt(x);
C:=expand(A*B)/B;
``(expand(combine(numer(C)/sqrt(x))))/map(p->expand(combine(p/sqrt(x))),denom(C)) assuming x>0;
                                

Using the above limits, we get the result  1/2 

For those who want to check the result in Maple (without any steps) , here's a procedure that allows you to automatically generate the expression  a[n]  for any  n :

restart;
a:=proc(n)
if n=1 then return sqrt(x) else
sqrt(x+a(n-1)) fi;
end proc:

Examples of use:

seq(a(n), n=[1,4,10]);

The answer is very simple - your inequality in addition to the basic unknown  r  contains several parameters  (lambda, M) and Maple simply does not know how to solve such inequalities.

Here is a much simpler example that Maple also refuses to solve:

solve(sqrt(a*x+1) < 1, x);
     Warning, solutions may have been lost

This simple example, we can still solve, if to use  assuming  option:

solve(sqrt(a*x+1) < 1, x) assuming a>0;
solve(sqrt(a*x+1) < 1, x) assuming a<0;

But for your inequality and it does not help.

I think first you need to specify the values of the parameters  lambda  and  M .

See  ?plot,axis  help page for this.

Do you just want to check that a number  a  is the root of your equation? In this case, do 

simplify(eval(F(X)=0, X=a));

Sometimes other commands are required instead of  simplify . Then specify the exact form of your equation and the value of a .
 

restart;
with(Physics):
Setup(mathematicalnotation = true);                
Setup(signature = `+---`);
ds2 := ((x^2-y^2)*cos(2*u)+2*x*y*sin(2*u))*(du^2)-2*x*(dv^2)-dx^2-dy^2;
Setup(coordinatesystems=(Z=[u, v, x, y]), metric = ds2);
 

Edit.

[{1, 2}, {2, 3}] minus~ [{2}$2];
                                                      [{1}, {3}]

A more traditional and programmatic way of obtaining the values of a function obtained as a result of solving an equation, etc., is to use  eval  command. See the same example as tomleslie's one:

Download eval.mw

See  question_(2)_new.mw

ode:= diff(y(x),x)=2*x:
plots:-display(DEtools:-DEplot(ode,y(x),x = -2 .. 2,y = -2 .. 2, [[0.1,0]],
               labels=["",""],
               linecolour = red,
               color = blue,
               'arrows' = 'medium',
               axesfont=['Times', 'bold', 12]
               ), axis=[tickmarks=[color=red]]);

For your first example, everything is the same:

restart;
plots:-display(plot(sin(x),x=-Pi..Pi, color=blue),axis=[tickmarks=[color=red]]);
 

Here is an example of solving the problem. The method is of a general nature and can be used for any curves given by parametric equations. In the example, the red curve is a broken line, and the blue one is a straight line.

restart;
d:=(P,Q)->sqrt((P[1]-Q[1])^2+(P[2]-Q[2])^2);
P:=piecewise(s>=0 and s<0.3,[s,s/3+0.3],s>=0.3 and s<0.5,[s,-s+0.7],s>=0.5 and s<0.7,[s,7/2*s-1.55],s>=0.7 and s<=1,[s,-s/3+17/15]);
Q:=[t,t];
A:=plot(piecewise(seq(op([op(P)[2*n-1],op(P)[2*n][2]]), n=1..nops([op(P)])/2)), s=0..1,color=red, thickness=3):
B:=plot(t, t=0..1, color=blue, thickness=3):
epsilon:=0.3;
C:=plots:-inequal({d(P,Q)<=epsilon}, s=0..1, t=0..1, optionsfeasible = [color = "WhiteSmoke"], optionsexcluded = [color = "DarkGrey"], view=[0..1,0..1]):
plots:-display(<plots:-display(A,B) | C>, scaling=constrained); # This is "free space diagram"

If the curve is given by a conventional equation, for example  y=f(x) , so you must specify it as the list  [x, f(x)]

FD.mw

Edit.

Try  
A.C;  # dot product 
# or  
expand(A.C); 

Type this ( . ) as an ordinary dot on the keyboard. I use 1d-math input. In 2d-math input (which you use) , this will look like a fat dot in the middle.

For real domain use  discont  command instead of  singular  (the output is empty set):

discont(ln(y^2+1), y);
                                           { }