Why is free radical chlorination exothermic whereas bromination is endothermic?
First off, we can see that both reactions are exothermic If we do the thermodynamic analysis.

Given that your starting materials have the same heat of formation: 

$$\Delta H^\circ_f(Products) = \Delta H^\circ_f (\ce{R-H}) + 0, \quad{\Delta H^\circ_f(\ce{Cl}) = \Delta H^\circ_f(\ce{Br})= 0}$$

$$\ce{R = Et}\\ \Delta H^\circ_f (\ce{EtH}) = \pu{-84 kJ mol^{-1}}$$
$$ 
\begin{array}{cccccc}
\hline
\text{Product 1} &amp; \Delta \text{H}^\circ_f [\pu{kJ mol^{-1}}] &amp; \text{Product 2} &amp; \Delta \text{H}^\circ_f [\pu{kJ mol^{-1}}] &amp; \Sigma\ \Delta \text{H}^\circ_f\  [\pu{kj mol^{-1}}] &amp;\text{Result}\\
\hline
\ce{HBr} &amp; -\text{36.45} &amp; \ce{Et-Br} &amp; -\text{97.6} &amp;-\text{134.05}&lt;-\text{84} &amp;\text{Exothermic} \\
\ce{HCl} &amp; -\text{92.31} &amp; \ce{Et-Cl} &amp; -\text{137} &amp; -\text{229.31}&lt;-\text{84}  &amp;\text{Exothermic}\\
\hline
\end{array}
$$



Back to your question:


  Why is there a difference?


The graphs you are looking at only consider the reaction:

$$\ce{R-H + X^. -&gt;  R^. + HX}$$

For which we get:
$$ 
\begin{array}{cccccc}
\hline
\text{X}
&amp; \Delta \text{H}^\circ_f(\ce{X^.})
&amp; \Delta \text{H}^\circ_f(\ce{RH})
&amp; \Delta \text{H}^\circ_f(\ce{R^.})
&amp; \Delta \text{H}^\circ_f(\ce{HX})
&amp; \Delta \text{H}^\circ_{rxn}
&amp; \text{Result}\\
\hline
\ce{Br} &amp; -\text{96.94}^{**} &amp; -\text{84} &amp; -\text{119}^\dagger &amp;-\text{36.45} &amp;\text{25.49}&amp;\text{Endothermic} \\
\ce{Cl} &amp; -\text{121.29}^{**} &amp; -\text{84} &amp; -\text{119}^\dagger &amp; -\text{92.31} &amp;-\text{6.02}&amp;\text{Exothermic}\\
\hline
\end{array}
$$
*all units in $[\pu{kj mol^{-1}}]$ all values from Wikipedia unless otherwise stated 
** Table of Bond Dissociation Energies
$^\dagger$ Values from NIST WebBook

Here the enthalpy of radicalization of the alkane is different for chlorine or bromine, but the halogenation in total is exothermic