FunctionRange, MinValue and MaxValue unable to give result

This can be done by the change Abs[y] -> a^2, Abs[z] -> b^2 in order to obtain a polynomial in a and b as follows.

Maximize[{1/
 32 (8 - Sqrt[(-8 - 36 Abs[y] (1 + 2 Sqrt[Abs[z]]) - 
       27 (Abs[y] + 2 Abs[y] Sqrt[Abs[z]])^2)^2 - 
    64 (1 + Abs[y] + 2 Abs[y] Sqrt[Abs[z]])] + 
  36 Abs[y] (1 + 2 Sqrt[Abs[z]]) +  27 (Abs[y] + 2 Abs[y] Sqrt[Abs[z]])^2), 
   0 < Abs[y] < 1 && 0 < Abs[z] < 1 && Sqrt[Abs[y]] > 0 && 
Sqrt[Abs[z]] > 0} /.{Sqrt[Abs[y]]-> a,Sqrt[Abs[z]] -> b,Abs[y]->a^2,Abs[z] -> b^2}, {a, b}]
(*{1/4, {a -> 0, b -> Indeterminate}}*)

and the warning (not an error)

Maximize::natt: The maximum is not attained at any point satisfying the given constraints.

The result means that b may be an arbitrary value (of course, 0<b&&b<1. Making use of Minimize instead of Maximize, one obtains

(*{1/32 (359 - 35 Sqrt[105]), {a -> 1, b -> 1}}*)

and a similar warning. It remains to notice that the polynomial, being a continuous function, takes each value between its minimum value and maximum value on a (compact) set a>=0&&a<=1&&b>=0&&b<=1. Numeric calculations with NMaximize and NMinimize with Method->"DifferentialEvolutionconfirm those results. Summarizing the above, one may conclude that the range under consideration is $(frac{1}{32} left(359-35 sqrt{105}right),frac 1 4)$.

expr = 1/32 (8 - 
     Sqrt[(-8 - 36 Abs[y] (1 + 2 Sqrt[Abs[z]]) - 
          27 (Abs[y] + 2 Abs[y] Sqrt[Abs[z]])^2)^2 - 
       64 (1 + Abs[y] + 2 Abs[y] Sqrt[Abs[z]])] + 
     36 Abs[y] (1 + 2 Sqrt[Abs[z]]) + 
     27 (Abs[y] + 2 Abs[y] Sqrt[Abs[z]])^2);

Looking at the plot,

Plot3D[expr, {y, -1, 1}, {z, -1, 1},
 PlotPoints -> 100,
 MaxRecursion -> 5,
 ClippingStyle -> None,
 AxesLabel -> (Style[#, 14, Bold] & /@ {y, z, t}),
 WorkingPrecision -> 20,
 PlotRange -> All]

max = expr /. y -> 0

(* 1/4 *)

min = Min[
  Limit[expr, #] & /@ {{y -> -1, z -> -1}, {y -> -1, z -> 1}, {y -> 1,
      z -> -1}, {y -> 1, z -> 1}}]

(* 1/32 (359 - 35 Sqrt[105]) *)

min // N

(* 0.0111476 *)