Replacing

There is one built-in function and one built-in statement you can use for replacing strings.

REPLACE(expression, find, replace [,start] [,count] [,compare])

You can use the REPLACE function to return a text string with a number of characters replaced.
This built-in function is case sensitive by default.

Replace("C:\Temp\", "C:\", "E:\") = "E:\Temp\" 
Replace("C:\Temp\", "C:\", "E:\", 1) = "E:\Temp\" 
Replace("c:\Temp\", "C:\", "E:\", 1) = "c:\Temp\" 

If you would prefer the function to not be case sensitive, you can change the default "compare" argument.
vbCompareMethod

Replace("c:\Temp\", "C:\", "E:\", 1, , vbCompareMethod.vbTextCompare) = "E:\Temp\" 

MID(stringvar, start [,length]) = string

You can use the MID - Statement to replace a specified number of characters with characters from another string.
Not to be confused with the MID - Function which returns a specified number of characters.

Dim sText As String 
sText = "C:\Temp\" 
Mid(sText, 1, 1) = "E" 
Debug.Print sText        '= "E:\Temp\"  

Dim sText As String 
sText = "C:\Temp\" 
Mid(sText, 4, 4) = "Personal" 
Debug.Print sText        '= "C:\Personal\"  

Replacing from the middle

You can use the INSTR function to return the position of a substring within a larger string, starting at the beginning.
You can use the LEFT function to return a number of characters from the left of a string.
You can use the RIGHT function to return a number of characters from the right of a string.

Dim sText As String 
Dim sNewText As String 
Dim istartchar As Integer 
Dim iendchar As Integer 
sText = "This is very long text" 
istartchar = InStr(sText, " is ") + Len(" is ") - 1 
iendchar = InStr(sText, " text") 

sNewText = Left(sText, istartchar) & _ 
           "quite short" & _ 
           Right(sText, Len(sText) - iendchar + 1) 
Debug.Print sNewText      '= "This is quite short text"  

Replacing the first occurrence

You can use the REPLACE function to return a text string with a number of characters replaced.

Dim sText As String 
sText = "The First Occurrence" 
Debug.Print Replace(sText, "e", "", 1, 1) 
'= "Th First Occurrence"

Dim sText As String 
sText = "AA BB CC BB AA" 
Debug.Print Replace(sText, "BB ", "", 1, 1) 
'= "AA CC BB AA"

Replacing the last occurrence

You can use the STRREVERSE function to return the text string with the characters reversed.

Dim sText As String 
sText = "The First Occurrence" 
Debug.Print StrReverse(Replace(StrReverse(sText), StrReverse("e"), StrReverse(""), 1, 1)) 
'= "The First Occurrenc"

Dim sText As String 
sText = "AA BB CC BB AA" 
Debug.Print StrReverse(Replace(StrReverse(sText), StrReverse("BB "), StrReverse(""), 1, 1)) 
'= "AA BB CC AA"

Replacing double quotes

You can use the REPLACE function to return a text string with a number of characters replaced.

Dim sText As String 
sText = "Replace ""double"" quotes" 
Debug.Print Replace(sText, Chr(34), "'")  
'= "Replace 'double' quotes"

Dim sText As String 
sText = "Replace ""double"" quotes" 
Debug.Print Replace(sText, """", "'")  
'= "Replace 'double' quotes"

Replacing line break

You can use the REPLACE function to return a text string with a number of characters replaced.

Dim sText As String 
sText = "first line" & vbCrLf & "second line" & vbCrLf & vbCrLf 
Debug.Print Replace(Replace(sText, Chr(10), ""), Chr(13), "") 
''= "first linesecond line"

Dim sText As String 
sText = "first line" & vbCrLf & "second line" & vbCrLf & vbCrLf 
Debug.Print Replace(sText, vbNewLine, "") 
''= "first linesecond line"