okay, I'll bite. They're both wrong.
The Action lab is specifically incorrect in conflating kinetics (reaction rate) with thermodynamics (equilibrium). Pretty much all reactions go faster with higher temp, but in the case of reversible exothermic reactions, the reverse reaction increases in rate more quickly than the forward reaction, so the final equilibrium state has more reactant remaining.
Where Action Lab is correct and The Chemist is wrong is in the application of Le Chatelier's principle to heat. The van't Hoff equation shows clearly that the change in the equilibrium constant for a reaction with temperature is dependent on the enthalpy change of the reaction (and nothing else):
$\ln {\frac {K_{2}}{K_{1}}}={\frac {\Delta _{r}H^{\ominus }}{R}}\left({\frac {1}{T_{1}}}-{\frac {1}{T_{2}}}\right)$
So if $\Delta_r H^\circ < 0$ and $T_2 > T_1$, then $K_2 < K_1$.
Where The Chemist is correct is when he points out that the rection can change as the temperature increases, which can affect this result. In the van't Hoff equation, we've assumed that $\Delta_r H^\circ$ is constant over the temperature range of interest. This isn't always true, but in most cases is a good approximation.
Strangely, after pointing this out, he makes exactly that mistake himself. In his counterexample of NaOH increasing in solubility with temperature, he wrongly assumes that the published enthalpy of solution of NaOH is the relevant value in his experiment. That value unfortunately is not applicable to this situation.
For any compound the "enthalpy of solution" is defined for the case of infinite dilution. So for NaOH, that means adding solid NaOH into pure neutral water in such a large volume that the added NaOH does not appreciably change the pH. In that case, there are two reactions going on:
(1) The separation of the NaOH lattice and hydration of $\ce{Na+}$ and $\ce{HO-}$ ions: $$\ce{NaOH(s) <=> Na+(aq) + HO-(aq)}$$
This reaction actually has a positive enthalpy change.
(2) The reaction of hydroxide ion with hydronium ions: $$\ce{HO- + H3O+ <=> 2H2O}$$
This reaction has a large negative enthalpy change.
In the case of infinite dilution (pH=7), both reactions are occurring, so the large negative enthalpy change of (2) results in a net negative enthalpy change, ie the reaction is exothermic.
However, when we get to the point that the solution is saturated, the pH is extremely high and there is very little hydronium left to react, so reaction (2) contributes much less, and the positive enthalpy change of reaction (1) dominates, making the reaction endothermic. (More accurately, the amount of hydronium is so low relative to hydroxide that the decrease in concentration of hydronium necessary to restore Kw for a given amount of added hydroxide is relatively very small.) Since the maximum solubility is determined by the behavior of a saturated solution, it is this enthalpy change that determines the change in solubility with temperature. That means that NaOH exhibits the expected behavior of an endothermic salt dissolution, ie increasing solubility with temperature.
