Rates of hydrolysis in bicyclic and monocyclic compounds
I’m going to shamelessly steal electronpusher’s drawing for clarification:



Two general hydrolysis mechanisms are possible: $\mathrm{S_N2}$ and $\mathrm{S_N1}$. First, let us consider $\mathrm{S_N2}$ since the analysis is more simple.

Both of the two species are tertiary bromides meaning that $\mathrm{S_N2}$ reactions are already severely inhibited. However, the second compound is even worse: if the nucleophile tries to attack from behind to interact with the $\sigma^*_{\ce{C-Br}}$ orbital, it would need to come from inside the bicyclo[2.2.2]octane system. This is completely impossible, not only due to sterics. Thus, while compound 1 is very slow in $\mathrm{S_N2}$ reactions, it is not possible for 2 to react in an $\mathrm{S_N2}$ manner.

Conclusion: 1 should react faster by $\mathrm{S_N2}$ — albeit still unmeasurably slow.



To undergo an $\mathrm{S_N1}$ reaction, first the halide must dissociate to give a carbocation. Both species are, as noted above, tertiary, meaning that carbocations are well stabilised by triple hyperconjugation. Initially, we expect both $\mathrm{S_N1}$ mechanisms to be fast.

A carbocation when formed must practically always adopt a planar, $\mathrm{sp^2}$ hybridised configuration. Thereby, the unoccupied orbital has the highest energy and all bonding electrons are as low as possible in energy. Any distortion away from the perfect $\mathrm{sp^2}$ geometry requires the empty orbital to be stabilised at the cost of populated orbitals — very unfavourable.

For bridgehead carbons as in 2, especially those that bridge small rings (which is the case here), the general rule is that they do not undergo reactions that require planar transition states for exactly that reason: they are too strained to adopt an $\mathrm{sp^2}$ configuration. (This configuration would not only require the bridgehead to ‘move inwards’, but would also strain the bonds of the three neighbouring carbons, whose angles would have to be decreased.)

Conclusion: while 1 reacts rapidly via $\mathrm{S_N1}$, compound 2 does not react via $\mathrm{S_N1}$.



Complete conclusion: 2 is unreactive towards hydrolysis. 1 therefore reacts much faster.