import lxml.html as lh
import urllib2
def text_tail(node):
    yield node.text
    yield node.tail
url='http://bit.ly/bf1T12'
doc=lh.parse(urllib2.urlopen(url))
for elt in doc.iter('td'):
    text=elt.text_content()
    if text.startswith('Additional  Info'):
        blurb=[text for node in elt.itersiblings('td')
               for subnode in node.iter()
               for text in text_tail(subnode) if text and text!=u'\xa0']
        break
print('\n'.join(blurb))
yields
  For over 65 years, Carl Stirn's Marine
  has been setting new standards of
  excellence and service for boating
  enjoyment.  Because we offer quality
  merchandise, caring, conscientious,
  sales and service, we have been able
  to make our customers our good
  friends.
  
  Our 26,000 sq. ft. facility includes a
  complete parts and accessories
  department, full service department
  (Merc. Premier dealer with 2 full time
  Mercruiser Master Tech's), and new,
  used, and brokerage sales.
Edit:  Here is an alternate solution based on Steven D. Majewski's xpath which addresses the OP's comment that the number of tags separating 'Additional  Info' from the blurb can be unknown:
import lxml.html as lh
import urllib2
url='http://bit.ly/bf1T12'
doc=lh.parse(urllib2.urlopen(url))
blurb=doc.xpath('//td[child::*[text()="Additional  Info"]]/following-sibling::td/text()')
blurb=[text for text in blurb if text != u'\xa0']
print('\n'.join(blurb))