Marcia,
Font color can be defined using a colorindex or just a long value or the three color components Red, Green, Blue (you just have to figure out what to pass). E.g., for your tenth element:
Code:
Sub ScratchMacro()
'A basic Word macro coded by Greg Maxey
Dim oPar As Paragraph
Dim oRng As Range
Dim arrParts() As String
For Each oPar In ActiveDocument.Range.Paragraphs
Select Case oPar.Range.ListFormat.ListLevelNumber And oPar.Range.ListFormat.ListString <> vbNullString
'Uses a standard ColorIndex value.
Case 1: PaintFont oPar.Range, 1
Case 2: PaintFont oPar.Range, 6
Case 3: PaintFont oPar.Range, 12
Case 4: PaintFont oPar.Range, 7
Case 5: PaintFont oPar.Range, 11
Case 6: PaintFont oPar.Range, 2
Case 7: PaintFont oPar.Range, 3
Case 8: PaintFont oPar.Range, 10
Case 9: PaintFont oPar.Range, 13
Case Else
arrParts = Split(oPar.Range.Text, Chr(9))
Select Case Len(arrParts(0))
Case 19
Set oRng = oPar.Range
oRng.MoveStart wdCharacter, 19
'Uses Red, Green, Blue color components.
PaintFont oRng, 0, 155, 233
End Select
End Select
Next
lbl_Exit:
Exit Sub
End Sub
Sub PaintFont(oRng As Range, lngColorIndex As Long, _
Optional lngG As Long = -1, Optional lngB As Long = -1)
If InStr(oRng.Text, ChrW(8211)) > 0 Then
oRng.MoveEndUntil ChrW(8211), wdBackward
oRng.End = oRng.End - 2
Else
oRng.End = oRng.End - 1
End If
If lngB = -1 Then
oRng.Font.ColorIndex = lngColorIndex
Else
oRng.Font.Color = RGB(lngColorIndex, lngG, lngB)
End If
End Sub