I have "somewhat" of a hypotetical question
If SF played another draw (yes extremely likely.....)
Would that give them a 9-3-2 record and basically be identical to a loss. or would it be counted as two half points making a full and go to 10-3
No, not quite.
It would take them to an equivalent record of 10-4. They would get 0.5 *win* for each tie, giving them the equivalent of 10 wins... but they have played 14 games at that point so you would have to divide the 10 wins by 14 total games to get a 0.714 winning %, which is what the NFL uses to determine Division winners and Wild Card winners.
Taking this potentially out to the end of the regular season...
If SF finishes 10-4-2 (going 1-1-1 in their last three games) and the Hawks finish 11-5 (thereby winning out), the teams would finish with identical winning %s, at 0.6875.
You would then have to start applying the tiebreaker rules, first for the Division Championship. First tiebreaker is Head-to-Head, which would be a wash assuming their one remaining loss would be to us.
The 2nd divisional tiebreaker is record within the division.
Assuming that SF ties Arizona in their last game of the year, their record within the division would be 2-2-2, which yields a 0.500 winning %. And Seattle would finish 3-3 and have the identical winning % with the division.
On to the next divisional tiebreaker... Best won-lost-tied percentage in common games (games played against the same opponents)... where it gets a bit hairy.
Common opponents for the 49-ers and Hawks outside of the NCFW are GB, Det, Minn, NYJ, Buff, Chicago, Miami and NE.
For SF to get to 10-4-2 and for Seattle to get to 11-5, and to also have the same winning % within the division, SF would have to lose to the Hawks and tie the Cardinals in Week 17... which means that they must necessarily beat NE next Sunday.
For the common eight opponents, Seattle would go 6-2 in this scenario... and SF would go 7-1, thereby winning the Division... unfortunately.
Hope that all makes sense.
My head hurts... no mas, por favor.