A comet orbits the Sun in the same plane as Earth's orbit
I'll just give an outline about how I might try to answer these questions, rather than answering them outright.
Answer the first question using the circular form of the Vis-Viva equation: $v=\sqrt{\frac{\mu}{r}}$, where $\mu$ is the standard gravitational parameter ($4\pi^2 AU^3/yr^2$) and $r$ is 1 AU.  If your answer isn't $2\pi$ AU per year, you probably did something wrong with the units.
For the second question, try to determine the shape of the comet's orbit using the inequality for escape velocity: $v_e \ge \sqrt{\frac{2\mu}{r}}$.  Once you have the orbit shape, the rest of the problem falls out pretty easily using known equations for that orbit.

 Escape velocity at 0.5 AU is
 $$v_e=\sqrt{\frac{2*4\pi^2}{0.5}}=\sqrt{16\pi^2}=4\pi$$

 which is the exact speed given in the question. That means the orbit
 is parabolic.  From  Bogan's table pulled from
 Bate/Mueller/White's Fundamentals of Astrodynamics, for a parabolic
 orbit, $$r=\frac{2q}{1+\cos{\theta}}$$ where $r$ is the distance from
 the central body, $q$ is the closest distance of the orbit, and
 $\theta$ is the angle from periapsis or true anomaly. Solving gives us
 $\theta=\pi/2$.

 The same table gives us the angle of velocity relative to the
 perpendicular to the radial direction: $$\phi = \theta/2 = \pi/4$$

 So the orientation of the comet's velocity is at
 $\pi/4$ or 45 degrees off a tangent line to a circle around the Sun
 with radius 1 AU as the comet passes through it.