Although you wrote F:=(x,y)->(x+y,2*x^2+y^2), I suspect that you would rather have F as a vector-valued function.  Additionally, since F is a function from R² to R², the proper name of the derivative is the Jacobian, not the gradient.  With these in mind, here is what we do:

Write a proc f which receives a guess for T and produces a new guess:

f := proc(T)
  local Tnew;
  blah blah ... compute the new T
  return Tnew;
end proc;

Then apply the proc in a loop, as in:

T1 := starting value of T;
T := f(T1);
while abs(T-T1)> 1e-3 do
  T1 := T;
  T := f(T1);
end do;

Upon exit from that loop, the value of T will be approximately what you expect.  The accuracy iis controlled by the value 1e-3 in the code above.  You may increase or decrease that as needed.

PS: I saw rlopez's note after I posted this.  And he is right; the way you have worded makes no sense.  In what I have shown I have assumed that we want two successive values of T to be the same, which agrees rlopez's guess.

PPS: In general there is no guarantee that such an iteration will converge.  You may want to add a counter which is incremented in each pass of the loop, and break out of the loop if the number of iterations exceeds a prescribed maximum.

Change the last line to:

plot( [seq(seq(eval(lambda, Nb=j), j in [0.1,0.2,0.3]), F in [0.1,0.2,0.5])], delta2=0.02..0.1);

Regarding your remrk "the movement seems to be impossible in real life", yes, that is impossible.  If you inspect closely the animation that you have posted, you will note that at certain point the tetrahedra penetrate each other.

However, it can be done with 8 tetrahedra without penetration.  Here it is:

Here is the worksheet that produced it: kaleidocycle.mw

And here is a variant where each tetrahedron is painted in four colors:

and here is the worksheet: kaleidocycle8-alt.mw

——————————————————

For whatever it's worth, here is the self-penetrating six-tetrahedron version:

and here is the corresponding worksheet: kaleidocycle6.mw

Posting several screenfuls of Maple code onto a web page is not particularly helpful and can discourage people from reading. Why not just upload your Maple worksheet instead?  In the window where you write your message for posting, note the big fat green arrow.  Click on it to upload.

I waded through what you have posted and have only a vague idea of what you are attempting to  do.  It appears that you wish to plot an uppercase letter "A" and then rotate it by some angle.  If so, then don't build the angle in the definition of A.  Just make a plain "A" in the normal upright position and name it something, let's say A.  To rotate A about a point [x,y] by angle alpha, do
plottools:-rotate(A, alpha, [x,y]);
That's all.

Let t be the angle that the vector makes relative to the positive x axis.  Let L be the vector's length.  Then the vector is given by

v := L*<cos(t), sin(t)>;

In answering your previous questions, Kitonum has shown you the proper syntax for defining position vectors.  Let's call them r1 and r2 in this case.  You need to do:
r1 := t -> 2*t^2*_i+16*_j+(10*t-12)*_k;
r2 := t -> (20-6*t)*_i+4*t^2*_j+2*t^2*_k;

Then you may verify that r1(2) and r2(2) are equal, that is, the airplanes collide at t=2.

The velocities are the derivatives of the position vectors.  They are computed via:

v1 := D(r1);
v2 := D(r2);

The velocities at the time of collision are given by v1(2) and v2(2).  You
should be able to calculate the angle between those vectors.

Additionally:

1.  Heed rlopez's advice on the proper use of "evaluate" versus "solve".
2.  You will receive better help if instead of posting code fragments, you upload a complete worksheet.  In the window where you edit your message before sending, observe the big fat green arrow.  Click on it to upload worksheet.

Download mw.mw

I have no idea what your code is meant to plot, but if you wish to shift the origin of an existing plot to x=1, y=2, here is how:
newplot := plottools:-translate(oldplot,1,2);

In line 13 from the bottom, delete the extra "end if".

My interpretation of the problem's statement is different from acer's and kitonum's.  I view the first equation as defining γ as the root xx of the equation on the right-hand side. Similarly, the second equation defines ψ as the root xx of the equation on the right-hand side.  With that in mind, here is how to proceed.

Download mw.mw

You have a set of equation SysEq which are linear in the variables dq.  The coefficient matrix is obtained through
J := LinearAlgebra:-GenerateMatrix(SysEq, dq)[1];

In your case J is an 8x11 matrix.  To display it, do:
evalm(J);

@assma Here is the code to do that calculation.  I assume that your 6.28 is actually meant to be 2π.  Change it to anything else as needed.

Download mw.mw