Try:
Sub ArrFndRep()
Dim ArrFR As Variant, i As Long
ArrFR = Array("  ", " ", "[^t]{2,}", "^t", "[ ^t]{1,}^13", "^p", "[^13]{2,}", "^p")
If UBound(ArrFR) Mod 2 = 0 Then
  MsgBox "Something's Wrong Genious"
  Exit Sub
End If
With ActiveDocument.Range.Find
  .ClearFormatting
  .Replacement.ClearFormatting
  .Forward = True
  .Format = False
  .MatchWildcards = True
  .Wrap = wdFindContinue
  For i = 0 To UBound(ArrFR) Step 2
    .Text = ArrFR(i)
    .Replacement.Text = ArrFR(i + 1)
    .Execute Replace:=wdReplaceAll
  Next
End With
End Sub
When using wildcards:
.MatchWholeWord = True
.MatchCase = False
are invalid, so there's no point having them. You also don't need ^0013 - ^13 will do - and you should NOT use ^13 (or ^0013) as the Replace expression; otherwise the affected paragraphs will become one, joined together by what appears as a ¶ character. Having " {2,}" as a Find expression with " " as the Replace expression will reduce all space sequences to 1, whereas your previous post indicated you wanted sequences of 2 and 3 spaces reduced by 1 space each (meaning a 3-space sequence would be reduced to 2 spaces, not to 1 space).