I use a trick to solve equation:

 2 Cos[FindSequenceFunction[
    Table[ArcCos[
       1/2 RSolveValue[{a[n] == Sqrt[2 + a[n - 1]], a[1] == Sqrt[2]}, 
         a[m], n]], {m, 1, 7}] // FullSimplify, n]]

(*2 Cos[2^(-1 - n) \[Pi]]*)

I noticed some related discussions here.

No way to find a solution ,because we can't find a closed-form solution from:

This sounds very frustrating. As you can see, the corresponding limit exists strictly, so it must be mathematically rigorously provable. However, the present computational algebraic system is incapable of dealing with such problems.

See here: "https://en.wikipedia.org/wiki/Viète%27s_formula"

I also mentioned this wiki page in my reply above.

For the other two, is it possible to evaluate the limit of product on the nested list, with the index approaching infinity?

I tried the method shown below based on the notes here, but failed:

In[26]:= Remove[f];
f[1] = Sqrt[2];
f[n_] := Sqrt[2 + f[n - 1]]
Limit[Product[1/2 f[i], {i, 1, n}], n -> Infinity]

During evaluation of In[26]:= $RecursionLimit::reclim2: Recursion depth of 1024 exceeded during evaluation of -1020+i.

Out[29]= 
\!\(\*UnderscriptBox[\(\[Limit]\), \(n \[Rule] \[Infinity]\)]\) \!\(
\*UnderoverscriptBox[\(\[Product]\), \(i = 1\), \(n\)]\(Hold[
\*FractionBox[\(f[i]\), \(2\)]]\)\)

Try:

  Block[{n = 500}(*is not a Infinity*), $MaxExtraPrecision = 1000; 
   N[Times @@ Table[1/2 Nest[Sqrt[2 + #] &, Sqrt[2], i], {i, 0, n}] - 2/
     Pi, 20]]

 (*6.1081960994840413729*10^-303*)

From the documentation of Product:

If the range of a product is finite, i is typically assigned a sequence of values, with f being evaluated for each one.

And also:

If a product cannot be carried out explicitly by multiplying a finite number of terms, Product will attempt to find a symbolic result. In this case, f is first evaluated symbolically.

So, the errors you're getting are because it can't symbolically evaluate your term expression once i is plugged in. If you used a finite bound, the i would be replaced with an actual number in each term, and then your term expression could be evaluated. If you pick a large enough bound, you will be within the standard accuracy anyway, so, do you really need an infinite product?

What is your objective? Are you just trying to see if Mathematica is smart enough to know these formulas?