Solution:
a. Set $\angle BAC = \alpha$ and $\angle ACD = x$. We have $AC = AB \cos \alpha$, $BC = AB \sin \alpha$ and it follows by the Sine theorem for $\triangle ADC$ and $\triangle BDC$ that

$$\begin{aligned} & \frac{AC^{2}}{AD+CD} + \frac{BC^{2}}{BD+CD} = \frac{AC}{\frac{AD}{AC} + \frac{CD}{AC}} + \frac{BC}{\frac{BD}{BC} + \frac{CD}{BC}} \\ & = \frac{AB \cos \alpha}{\frac{\sin x}{\sin (\alpha + x)} + \frac{\sin \alpha}{\sin (\alpha + x)}} + \frac{\cos x}{\sin (\alpha + x)} + \frac{\cos \alpha}{\sin (\alpha + x)} \\ & = AB \left( \frac{\cos \alpha \sin (\alpha + x)}{\sin \alpha + \sin x} + \frac{\sin \alpha \sin (\alpha + x)}{\cos \alpha + \cos x} \right) \\ & = AB \left( \frac{\cos \alpha \cos \frac{\alpha + x}{2}}{\cos \frac{\alpha - x}{2}} + \frac{\sin \alpha \sin \frac{\alpha + x}{2}}{\cos \frac{\alpha - x}{2}} \right) = AB. \end{aligned}$$

The above identity can be proved by using the Stewart theorem as well.

b. We have
$$\frac{AE}{CE} = \frac{AD}{CD} \Rightarrow \frac{AE + CE}{CE} = \frac{AD + CD}{CD} \Rightarrow CE = \frac{AC \cdot CD}{AD + CD}$$
and similarly $CF = \frac{BC \cdot CD}{BD + CD}$. Then
$$\begin{aligned} \frac{CF}{CA} + \frac{CE}{CB} & = \frac{BC \cdot CD}{CA(BD + CD)} + \frac{AC \cdot CD}{CB(AD + CD)} \\ & = \frac{CD}{CA \cdot CB} \left( \frac{CA^{2}}{AD + CD} + \frac{BC^{2}}{BD + CD} \right) \end{aligned}$$
and using a) we see that
$$\frac{CF}{CA} + \frac{CE}{CB} = \frac{CD \cdot AB}{CA \cdot CB}$$
Therefore the required minimum is equal to $1$ and it is attained when $CD$ is the altitude of $\triangle ABC$.