LDA will not attach, but it may be strong enough to form an aryne. For example one could deprotonate C-H and elimination of the best leaving group (Cl) would give an aryne, which could then undergo attack by a nucleophile. When the nucleophile does attack, it will likely do so such that the negative charge is placed closer to the electron withdrawing fluoro groups.
See this artice on arynes – possibly helpful. https://www.masterorganicchemistry.com/2018/09/17/nucleophilic-aromatic-substitution-2-benzyne/

Sometimes on benzene rings with a good leaving group it will deprotonate C-H adjacent to the carbon with a good leaving group, resulting in a new pi bond. These species are called “arynes”. https://www.masterorganicchemistry.com/2018/09/17/nucleophilic-aromatic-substitution-2-benzyne/

Alkylation of enolates with alkyl halides is an SN2 reaction, and SN2 reactions will only occur at sp3-hybridized carbons. In the reaction you describe, only the 1-carbon is sp3-hybridized, and thus the enolate will perform substitution at C-1. The C-Br bond at C2 will be unaffected. Sorry for taking so long to get back to you.

No, but there will be a change in selectivity. It won’t be as selective for the less substituted enolate at a higher temperature.

Confusingly, the name for the amide linkage you speak of, and the name of the conjugate base of an amine are both “amide”. Some chemists pronounce the two differently but that doesn’t really translate to the written word. It’s helpful to use the term “metal amide” when referring to the conjugate base – this differentiates it from the (neutral) functional group.

LDA formation would essentially be instantaneous at –20 °C. In fact it’s not uncommon to make the LDA at sub-zero temperature and then cool it to -78 prior to using it for enolate formation.

The pKa of the CH3 adjacent to a triple bond is at least 40, so without significant heating, no.