Sia

Sure, let's analyze the given reactions and predict the products.

 Reaction (a)
The first reaction involves a six-membered cyclic structure with an SH group reacting with $CH_3O^-Na^+$. This is an acid-base reaction where the SH group (thiol) acts as an acid and $CH_3O^-$ (methoxide) acts as a base.

The reaction can be represented as:
$$\text{Cyclic structure-SH} + CH_3O^-Na^+ \rightarrow \text{Cyclic structure-S}^- + CH_3OH$$

The thiol group donates a proton to the methoxide ion, forming a thiolate anion and methanol.

 Reaction (b)
The second reaction involves a 12-membered cyclic structure with three sulfur atoms undergoing reduction. Reduction typically involves the gain of electrons or hydrogen atoms.

Without specific details on the reducing agent, a common reduction reaction for sulfur-containing compounds is the reduction of disulfide bonds ($S-S$) to thiols ($SH$).

The reaction can be represented as:
$$\text{Cyclic structure with } S-S \text{ bonds} \rightarrow \text{Cyclic structure with } SH \text{ groups}$$

 Position of Equilibrium
For the acid-base reaction in (a), the position of equilibrium can be predicted based on the relative strengths of the acids and bases involved. Methanol ($CH_3OH$) is a weaker acid compared to the thiol group, so the equilibrium will favor the formation of methanol and the thiolate anion.

 Answer Format
For the given questions:

(a) 

(b)