With these calculations you get lambda and cf as functions of a,pd,pf plus a compatibility condition between a,pd,pf:

(solLa = Solve[eqns1[[1 ;; 2]] /. lambda^a -> la, {la, cf}]) /. 
 la -> lambda^a
solLambda1 = 
 FullSimplify[
  Solve[Simplify[
    eqns1[[3]] /.
      { lambda^a -> la, 
       lambda^(1 - a) -> lambda/la} /. solLa[[1]]], lambda]]
solLambda2 = 
 FullSimplify[
  Solve[Simplify[
    eqns1[[3]] /.
      { lambda^a -> la, 
       lambda^(1 - a) -> lambda/la} /. solLa[[2]]], lambda]]
(la /. solLa[[1]]) == (lambda /. solLambda1)^a
(la /. solLa[[2]]) == (lambda /. solLambda2)^a

The compatibility condition a,pd,pf is not algebraic, because the variable a appears at the exponent. I doubt you can solve it explicitly.

Dear Hans, thank you very much for your message and work. What I am looking for is mainly to be able to compare pd and pf in function of a and cf (or their ranges). I admit this way seems very complicated... I have now solved for lambda using Gianluca's way, but Mathematica is still not able to solve for pd and pf the system made of the two first equations in a reasonable amount of time. I will let you know if I find some other easier way to do it...

Dear Olimipia, thanks for your answer.

But what exactly do you want? Look at the code below. Mma solved the 1st equation quite quickly for pd, but you get three (already complicated) solutions, and each should be inserted to the remaining equations to be solved further. The result is not really convenient.

Of couse one can solve the 1st equation for pf, giving only two solutions to be inserted in the rest, but is this easier? Not really.

f[x_] := x^a
fl = f[lambda];
dfl = D[f[lambda], lambda];
cd = 12/10;

pn = ((pf*(1 + fl))^(-2) + (pd^(-2)))^(-1/2);
sf = (pf*(1 + fl)/pn)^(-2);
sd = (pd/pn)^(-2);

eqns = {pd == (3*(1 - sd) + 6)/(3*(1 - sd) + 5)*cd, 
  pf == (3*(1 - sf*(1 + fl)^(-2)) + 
       2*sf*(1 + fl)^(-2))/(3*(1 - sf*(1 + fl)^(-2)) + 
       2*sf*(1 + fl)^(-2) - 1)*cf, (1 + fl)^3/dfl == 
   10*pn^3*pd^(-3)*(pd - cd)*pf^(-2)}


eq11 = eqns[[1]] // Simplify
solpd = Solve[eq11, pd]


solpf = Solve[eq11, pf]

Dear Hans, thank you for your message. I have the hope that I will be able to interpret in some general ways the effect of the parameters. As I said, the model is easy to solve numerically (even in a much more complex version) but my choices of parameters remain very arbitrary...

Hello Olimpia, why do you want to have a symbolic solution?

If it exists at all it will be terribly complicated ( I got intermediate expressions of third order and accordingly three solutions, some with complex terms) and I suppose you won't "see" anything with it.

Thank you for your answer. I understand your argument, but if I try to declare a=2/3 and let cf unknown it shouldn't be the case anymore. The three (simplified) equations to solve become

eqns={
5 pd + 18/5 (-2 + 1/(8 + (5 (1 + lambda^(2/3))^2 pf^2)/pd^2)) == 0, 
 pf + cf (-(3/2) + 1/(-2 (1 + 4 lambda^(2/3) + 2 lambda^(4/3)) - (4 (1 + lambda^(2/3))^4 pf^2)/pd^2)) == 0, 
 3/2 (1 + lambda^(2/3))^3 lambda^(1/3) == (10 (-(6/5) + pd))/(pd^3 (1/pd^2 + 1/((1 + lambda^(2/3))^2 pf^2))^(3/2) pf^2)
} 

but still, it seems Mathematica keeps running without finding a solution...