Add a new line or space depending on pattern - regex

Question

I am trying to do the following.
Patterns:
aaaaa.BBBBB - to add New Line after the (.)
aaaaaBBBBB - to add New Line when see a Caps letter.
aaaaa12345 - to add a space when there is a digit (Output: aaaaa 12345)
12345aaaaa - to add a space when there is a letter after the digit (Output: 12345 aaaaa)
Values:
Client asked about the 21year planPlease follow up at1234567
ReGex code need to the following:
Client asked about the 21 (space) year plan**(new line)** Please
follow up at (space) 1234567
Result:
Client asked about the 21 year plan
Please follow up at 1234567.
How do I recognize the pattern and also do a specific replacement be it adding (space) or (newline)?
Here is the code I use currently:
Function SplitCaps(strIn As String) As String
Dim objRegex As Object
Set objRegex = CreateObject("vbscript.regexp")
With objRegex
.Global = True
.Pattern = "([a-z])([A-Z0-9])"
SplitCaps = .Replace(strIn, "$1 $2")
End With
End Function

Answer 1

You can use two regex replacements. The first one to add spaces between 0a and a0 (between a number and a lowercase letter), and a second to create the newline between aA and a.A.
([a-z])([0-9])|([0-9])([a-z]) and replace with $1$3 $2$4
([a-z])\.?([A-Z]) and replace with $1\n$2
If you want a period added at the end use $ and replace with \.
Try this code:
Function SplitCaps(strIn As String) As String
Dim objRegex As Object
Set objRegex = CreateObject("vbscript.regexp")
Dim result As String
With objRegex
.Global = True
.Pattern = "([a-z])([0-9])|([0-9])([a-z])"
result = .Replace(strIn, "$1$3 $2$4")
End With
With objRegex
.Global = True
.Pattern = "([a-z])\.?([A-Z])"
result = .Replace(result, "$1\n$2")
End With
SplitCaps = result
End Function