Dissociation rates: trans-effect rule in square planar complexes
First off, there are two nice articles on the trans effect on Wikipedia and LibreTexts that are worth reading.

You need to remember that there are two factors leading to the trans effect:


Weakening the $\ce{Pt-X}$ bond trans to another group (i.e., a structural effect)
Stabilizing the 5-coordinate transition state, since substitution reactions in square planar complexes are generally associative. (i.e., a kinetic effect)


(a) I don't think the $\ce{I-}$ will substitute for the $\ce{NH3}$ group. Instead, I think the $\ce{CO}$ "wins" the trans effect "competition" and so the trans $\ce{CH3-}$ will be replaced by the $\ce{I-}$.

Next, substitution with the $\ce{Cl-}$ should replace the $\ce{I-}$.



Now your second question seems to be if you'll actually get a substitution with $\ce{Cl-}$ since $\ce{I-}$ has a greater trans effect. Yes, you will. All of these are in equilibrium, so if you generate a lot of $\ce{I-}$ that can substitute back (and get starting material). But you'll definitely get the Cl-substituted product too if you separate the reaction mixture.

Indeed, the prototypical example for the trans effect is synthesis of trans-platin or cis-platin, starting from either $\ce{[PtCl4]^{-2}}$ or $\ce{[Pt(NH3)4]^{+2}}$. Even though $\ce{NH3}$ has a smaller trans effect, you can synthesize cis-platin from $\ce{[PtCl4]^{-2}}$ and substituting two $\ce{Cl-}$ with $\ce{NH3}$.