I'm trying to calculate
(1): the probability of $HBP$ given $HD$ and $FH$, $P(HBP | HD, FH)$.
(2): probability of $SM$ given $HD$, $P(SM | HD)$.
I've calculated $P(HBP) \approx .468, P(HD) \approx .662$.
For (1) I'm not positive in how to go about this. The only thing I'm sure of is that $P(HBP | HD, FH) \neq P(HBP | HD)$ because $HBP$ and $FH$ are initially independent.
For (2) I have that $P(SM | HD) = \frac{P(HD|SM)P(SM)}{P(HD)}$, where, \begin{equation*} P(SM|HD) = \frac{P(HD|SM)(.20)}{(.662)} \end{equation*}
For $P(HD|SM)$ I believe I should calculate the $P(HD)$, but only when $P(HBP)$ is included (Y), since that includes $P(SM)$?
Since the calculations are kind of tedious, I am going to give you just a hint to the first one (not in comments, as I could not fit it). $$P(HBP|HD,FH)=\frac{P(HD|HBP,FH)P(HBP,FH)}{P(HD,FH)}$$ This could be simplified further to: $$P(HBP|HD,FH)=\frac{P(HD|HBP,FH)P(HBP|FH)P(FH)}{P(HD|FH)P(FH)},$$ and $FH$ cancels out. If $HD,FH$ are indeed independent $P(HBP|FH)=P(HBP)$.
$P(HD|HBP,FH)$ could be calculated from the table you provided, as $$P(HD|HBP,FH)=P(HD|HBP,FH,Ath)P(Ath)+P(HD|HBP,FH,NoAth)P(NoAth).$$ In a similar (but longer) manner $P(HD|FH)$ is a weighted sum of 4 rows from the table, anywhere, where FH is marked as "Yes".
Let me know, if it clarifies the matter.
As for the second issue: as I understand the picture, $SM$ causes $HD$ only if it causes $HBP$. I do not know if it is true in reality, but in the scheme it looks so.