Nancy,
Actually, you can accomplish this using a standard "XML" webpart and the "RSS" feed from the discussion list.
First, go to the "RSS" feed for the list ("Actions > View RSS Feed") and copy the web address (should be something like "http://<your portal>/_layouts/listfeed.aspx?List=<html-encoded Guid of the list>).
Next, add an "XML" webpart onto the page you want to view the most recent post's details on.
In the webpart's toolpane you'll see a textbox just under the top button ("XML Editor") where you can paste in the url you copied (textbox is called "XML Link").
Finally, you'll need to add in some custom XSLT to format the resulting XML and trim down the details to show only the most recent post, the total count for the specific thread, and who it was that posted.
If you're not too familiar with XSLT you can use the following as a sample to get you started:
<?xml version="1.0" encoding="utf-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="/">
<h2>Recent Posting</h2>
Post Title: <a href="{rss/channel/item/link}">
<strong>
<xsl:value-of select="descendant::rss/channel/item/title[1]"/>
</strong>
</a>
<br />
<xsl:for-each select="rss/channel/item[position() < 2]">
Number of Replies: <xsl:value-of select="count(//title[.=current()/title])" />
<br />
</xsl:for-each>
<br />
Posted by: <xsl:value-of select="descendant::rss/channel/item/author[1]"/>
<br />
On: <xsl:value-of select="descendant::rss/channel/item/pubDate[1]"/>
</xsl:template>
</xsl:stylesheet>
Hopefully this will get you going in the right direction.
- Dessie
