I'm pretty sure how to find arc length using the formula. But I'm stuck on 1a. I solved for x for the second equation and got $ \ x \ = \ \frac{(2-z)}{\sqrt{3}} \ \ . $ I then substituted this into the first equation to get $ \ \frac{4(2-z)^2}{3} + y^2 = 4 \ \ . $ Where do I go next? How do I parametrize and find the direction of the curve?
You've got $4x^2 + y^2 = 4$, so trig functions are probably a good place to start. Set $$\vec{r}(t) = (\cos(t) , 2 \sin (t), z(t))$$ One easily checks that $4x^2 + y^2 = 4$ is satisfied by this parametrization. It remains to choose $z(t)$ so that $\sqrt{3}x + z = 2$ is satisfied. But this is easy enough. Just set $z(t) = 2 - \sqrt{3}x(t) = 2 - \sqrt{3}\cos(t)$. We finally obtain the parametrization $$ \vec{r}(t) = (\cos(t) , 2 \sin (t), 2 - \sqrt{3} \cos(t)) $$ Finish up by calculating arc length using the standard integral formula.