Role of intermediate in determining the exothermic or endothermic nature of reaction
A typical method for determining whether a reaction is endo or exothermic is to compare the heats of formation ($\Delta_f H$) of the products to those of the reactants. If we do not have measured values for the enthalpies of formation, we can get a fairly good estimate by summing up typical values for the bond enthalpies of the bonds in the compounds.
When using bond enthalpies, we can simplify the calculation by considering only those bonds that are broken or formed during the reaction, since the other bonds appear in both the products and reactants and so cancel out when we look at the difference.
Thus,
$$\textrm{Enthalpy of reaction = }\Delta_r H = \textrm{[sum of enthalpies of bonds broken]} - \textrm{[sum of enthalpies of bonds formed]}$$
[Here I'm using the standard convention that bond enthalpies are reported for the breaking of bonds and so are positive values.]
Comparing the four reactions given, we can simplify even further by noting that all four reactions involve breaking of a $\ce{H3C-Cl}$ bond and formation of a $\ce{H3C-H}$ bond. Since those occur in all four reactions, we can ignore them for the purpose of comparison.
We now have reduced the problem to a simple comparison of the change of an $\ce{R-H}$ bond to an $\ce{R-Cl}$ bond.
You are correct that it does not matter what intermediates might occur on the reaction path. The radical stability is relevent, however, because the bond enthalpy is defined as the change in enthalpy needed for homolytic bond cleavage, ie
$\ce{A-B -&gt; A. + B.}$
Since the products are radicals, the stability of the radicals gives us information about the bond enthalpy. It is for this reason that comparing the stability of radical &quot;intermediates&quot; is a way to tell which of your given reactions is not exothermic.