Number 2 is really amazing. I thought I saw it sometimes somewhere, and even a proof, but I did not find it again. It seems convergence is really fast.

A code to get the expression

nn = 3;
ppp = Join[{Sqrt[2]}, 
  hh[ Table[(z = If[OddQ[k], k + 1, k])/ If[OddQ[k], k, k + 1], #]]^(1/(2 #[[2]])) & /@ 
   Table[{k, 2^j, 2^(j + 1) - 1}, {j, 1, nn}]]
Times @@ (ppp /. hh[x_] :> Times @@ x) // N
E/2 // N

or to check convergence

ff[n_] := Module[{},
  ppp = Join[{Sqrt[2]}, 
    hh[ Table[(z = If[OddQ[k], k + 1, k])/ If[OddQ[k], k, k + 1], #]]^(1/(2 #[[2]])) & /@ 
     Table[{k, 2^j, 2^(j + 1) - 1}, {j, 1, n}]];

  Times @@ (ppp /. hh[x_] :> Times @@ x) // N
  ]

ListLinePlot[
 Table[{j, ff[j]}, {j, 2, 6}],
 Epilog -> {Red, Line[{{2, E/2}, {6, E/2}}]},
 PlotRange -> {1.35, 1.37}
 ]

BTW, how to solve the original question 2?

From here:https://mathworld.wolfram.com/PippengerProduct.html

   Sqrt[2]*Product[((
       2^(-1 + 2^n)
         Gamma[1/2 + 2^(-2 + n)]^4)/(\[Pi] Gamma[
         1/2 + 2^(-1 + n)]^2))^(1/2^n), {n, 2, Infinity}]

  (*E/2*)

Finally, the problem becomes the following: How to find its equivalent closed form which can be calculated analytically?

This site: https://math.stackexchange.com is more appropriate.

What is the idea behind this technique, and how did you come up with this method?

The art and craft of it---which only comes with practice---is having the intuition to guess when it's a good time to try that.

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]]*)

.

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[\(f[i]\), \(2\)]]\)\)

Why can equation 3 in my example be analytically verified perfectly, while the other two cannot be calculated similarly?

I want to use any feasible calculation method, say, taking the Limit to accurately verify the validity of those equations with the help of Mathematica.