Why is a C–D bond stronger than a C–H bond?
Physics is better able to answer "how" questions than "why" questions, but here goes. The quantum mechanical description of the C–D system versus the C–H system gives the former a lower zero-point energy, which is the minimum energy the quantum system can attain. A good conceptual model of this is to consider the C–X system as two masses connected by a spring. When X = D, the system will vibrate more slowly than when X = H simply due to mass.

Since energy is proportional to the frequency of vibration, the energy of the C–D system is less (lower). This table lists the C–D bond dissociation energy as $\pu{341.4 kJ/mol}$ and C–H as $\pu{338 kJ/mol}$. Since this is the energy to break the bond, the C–D bond is stronger.



As far as your question about hyperconjugation; I'm surprised that C–H would have greater hyperconjugation than C–D, but it is not something that I know anything about. I guess I could speculate and wave my hands around and say that since the C–D bond is stronger, it keeps the electrons "closer" and so more confined to the sigma orbital, but that is just hot air.

Here's something which claims that the reason D does less hyperconjugation is because it vibrates less (see slide 15). So, it seems (if this is correct) that it's the smaller bond deformations that reduce the hyperconjugation for deuterated systems. Here is a pretty clear explanation of the kinetic isotope effect.