Problem 62

Solution:

(1) When $a = \frac{1}{2}$, by combining $l$ and $C$ and eliminating $y$, we get $3x^2 + 2x - 2 = 0$.

Therefore, the length of the chord formed by the intersection of $l$ and $C$ is $\sqrt{2} \cdot \sqrt{\left(-\frac{2}{3}\right)^2 + \frac{8}{3}} = \boxed{\frac{2\sqrt{14}}{3}}$.

(2) By combining the line $l: x+y=1$ and the hyperbola $C: \frac{x^2}{a^2} - y^2 = 1$ and eliminating $y$, and after rearranging, we get $(1-a^2)x^2 + 2a^2x - 2a^2 = 0$.

Therefore, $\begin{cases} 1-a^2 \neq 0 \\ 4a^4 + 8a^2(1-a^2) > 0 \end{cases}$, solving this yields $0 \frac{\sqrt{6}}{2}$, and $e \neq \sqrt{2}$.

Hence, the range of values for the eccentricity $e$ of the hyperbola $C$ is $\boxed{\left(\frac{\sqrt{6}}{2}, \sqrt{2}\right) \cup \left(\sqrt{2}, +\infty\right)}$.