Problem 3: The key is to observe that \left(|\mathbf{r}(t)|^2\right)'=2\mathbf{r}(t)\cdot \mathbf{r}'(t).
Problem 5: Many students misunderstood the question. Since the engine is shut off at time t=0, My Skywalker is going to have a linear motion. His initial position is determined by \mathbf{r}(0) and his velocity is just \mathbf{r}'(0). Thus to determine whether he can make to the station is to check if the vector between his initial position and the station is parallel to his velocity.
Problem 6: Some students found the distance from the fly’s initial point to the point where it hits the sphere. But here what we really want to find is the arc length between those two points along the curve.
Problem 10: The first thing to observe is that the curve is actually a line even it does not look like one. Since x(t)=t^3, y(t)=2t^3, z(t)=-t^3, we always have 2x=y=-2z which is exactly the symmetric equation for a line. Once notice this, the rest is to parametrize the line (forget about the old parametric equation of t) such that the speed is 2. Note this line goes through the origin and its direction is (1,2,-1). In other words, the ‘velocity’ now is (1,2,-1) and we want to find a new ‘velocity’ such that its magnitude, i.e., speed, is 2. So the new ‘velocity’ is \frac{2}{\sqrt{6}}(1,2,-1). Thus the new parametric equation is r(s)=\frac{2s}{\sqrt{6}}(1,2,-1).