a. We only provide a proof for the configuration as shown. The other cases are similar.
Let $X$ and $Y$ be any points on the tangent at $R$ to $(ABR)$ and the tangent at $Q$ to $(ABQ)$ respectively as shown. We have
$$\angle MBR = \angle MAR = \angle YQA = \angle QBA,$$
and so
$$\begin{aligned} \angle XRA &= \angle RBA = \angle RBQ + \angle QBA = \angle RBQ + \angle MBR \\ &= \angle MBQ = \angle KAQ = \angle KAR. \end{aligned}$$
This implies $RX \parallel KA$.

b. Let the tangent at $R$ to $(ABR)$ and the tangent at $Q$ to $(ABQ)$ meet at $S$. Since
$$\angle BRS = \angle BAR = \angle BAQ = \angle BQS,$$
the points $Q$, $B$, $S$, $R$ are concyclic. This implies
$$\angle SBR = \angle SQR = \angle MBR,$$
and hence $S$, $M$, $B$ are collinear. Thus, the two tangents and $KM$ are concurrent.