a) We will prove that $SM$ and $SN$ are isogonal in $\angle ASB$, since $(SMN)$ touches $(SAB)$ and $J$ belongs to the line connecting $S$ and the center of $(SAB)$. Indeed, according to Steiner's theorem for pairs of isogonals, we need to show that
$$\frac{MA}{MB} \cdot \frac{NA}{NB} = \frac{SA^2}{SB^2}.$$
On the other hand, since $HM$ and $KN$ are symmedian of triangles $HAB$ and $KAB$, let $Z$ be the intersection of $HK$ with $AB$. The left hand side of the above equation can be calculated by
$$\frac{MA}{MB} \cdot \frac{NA}{NB} = \frac{HA^2}{HB^2} \cdot \frac{KA^2}{KB^2} = \frac{ZA^2}{ZB^2} = \frac{AF^2}{BE^2}.$$
Next, suppose that $SR, AR, BR$ meet $AB, RB, RA$ at $R', F', E'$ respectively. According to Thales's theorem and Ceva's theorem, we have
$$\frac{AF}{BE} = \frac{AF}{SR} \cdot \frac{SR}{BE} = \frac{AF'}{SF'} \cdot \frac{BE'}{SE'} = \frac{AR'}{BR'} = \frac{SA}{SB}.$$
In short, we get
$$\frac{MA}{MB} \cdot \frac{NA}{NB} = \frac{AF^2}{BE^2} = \frac{SA^2}{SB^2}.$$
so $SM, SN$ is isogonal in $\angle ASB$ and the center of $(SMN)$ lies on the fixed line.

b) By the lemma in Problem 3, $SE$ and $SF$ are isogonal with respect to $\angle ASB$. We will prove that the median at $G$ of the triangle $GPQ$ passing through the fixed point $L$ is the midpoint of the arc $CD$ that does not contain $S$ of $(SCD)$. Rewriting the problem in a more compact form as follows:
Let $SAB$ be a triangle with $I$ is the midpoint $AB$, any two points $C, D$ on $SB, SA$. Two points $U, V$ belong to the circumcircle of triangle $SAB$ such that $SU, SV$ is isogonal in $\angle ASB$ and $G$ is the intersection of $AU$ with $BV$. Let $d$ be the angle bisector $\angle ASB$, on the line through $B$ and $A$ parallel to $d$, take the points $P$ and $Q$ satisfying $CP \parallel SU$, and $DQ \parallel SV$. Let $T$ be the midpoint of $PQ$. Prove that $GT$ passes through the midpoint $L$ of arc $CD$ that does not contain $S$ of the circumcircle of triangle $SCD$.
Let $K$ be the second intersection of $SL$ and the circumcircle of triangle $SAB$, let $J$ be the second intersection of the circumcircle of triangle $SCD$ with the circumcircle of triangle $SAB$. It is easy to see that $TI$ is the midline of the trapezoid $AQPB$, so $TI \parallel AQ \parallel SL$. Therefore, by Thales's theorem, we only need to prove that
$$\frac{TI}{LK} = \frac{GI}{GK}.$$

First of all, we have
$$\begin{aligned} \frac{GI}{GK} &= \frac{GI}{GA} \cdot \frac{GA}{GK} = \frac{\sin GKA}{\sin GAK} \cdot \sin GAI \\ &= \frac{\sin UAB}{\sin UAK} \cdot \sin GKA = \frac{UB}{UK} \cdot \sin GKA. \end{aligned}$$
On the other hand, since
$$\angle QAD = \angle KSA = \angle AVK \text{ and } \angle QDA = \angle VSA = \angle VKA$$
then $\triangle BUK = \triangle AVK \sim \triangle QAD$ and similarly they are similar to $\triangle PBC$. On the other hand, by the rotation predicate we have
$$\triangle SAD \sim \triangle SKL \sim \triangle SBC.$$
From the above pairs of similar triangles, we have the ratio transformation
$$\frac{UB}{UK} = \frac{AQ}{AD} = \frac{BP}{BC} = \frac{AQ + BP}{AD + BC}.$$
Finally, since $AQ + BP = 2LK$, according to the property of the midline of the trapezoid and the Ptolemy's theorem,
$$KA(JA + JB) = JK \cdot AB$$
so we get the following
$$\begin{align*} \frac{TI}{LK} &= \frac{AQ + BP}{2LK} \\ &= \frac{AD + BC}{2LK} \cdot \frac{UB}{UK} \\ &= \frac{JA + JB}{2JK} \cdot \frac{UB}{UK} \\ &= \frac{AB}{2KA} \cdot \frac{UB}{UK} \\ &= \frac{UB}{UK} \cdot \sin GKA \\ &= \frac{GI}{GK}. \end{align*}$$
So the equality is proved and $GT$ passes through $L$. Therefore, the median of vertex $G$ in triangle $GPQ$ passes through the midpoint of arc $CD$ which does not contain $S$ of the circumcircle of triangle $SCD$, which is a fixed point. $\square$