The solution is based on the following
Theorem. Let $P = (x_0, y_0, 0)$ be a point in the xy-plane and $(a, b, 1)$ a vector perpendicular to the vector $v = (x_0, y_0, 0)$, that is, such that $a x_0 + b y_0 = 0$.
Then the set of points obtained by rotating the line $P + t \cdot v$, $t \in \mathbb{R}$, around the $z$-axis is the hyperboloid
$$x^2 + y^2 - (a^2 + b^2)z^2 = x_0^2 + y_0^2.$$
Moreover, each point of the hyperboloid belong to exactly one of the rotated lines.
Proof. A generic point in the line $\ell: P + t \cdot v$ is $Q = (x_0 + a t, y_0 + b t, t)$, $t \in \mathbb{R}$. Since
$$(x_0 + a t)^2 + (y_0 + b t)^2 - (a^2 + b^2) t^2 \\ = x_0^2 + y_0^2 + 2(a x_0 + b y_0)t + a^2 t^2 + b^2 t^2 - (a^2 + b^2)t^2 = x_0^2 + y_0^2,$$
$Q$ belongs to the aforementioned hyperboloid. Since rotating the line around the $z$-axis does not change the $z$-coordinate of $Q$ and the section of the hyperboloid in the plane $z = t$ is a circle, the image of $Q$ obtained by rotating $\ell$ belongs to the hyperboloid as well.
Conversely, if $(x_1, y_1, z_1)$ belongs to the hyperboloid, then it belongs to the plane $z = z_1$, whose section in the hyperboloid has equation $x^2 + y^2 = (a^2 + b^2)z_1^2 + x_0^2 + y_0^2 = (a^2 + b^2)z_1^2 + x_0^2 + y_0^2 + 2a x_0 z_1 + 2b y_0 z_1 = (x_0 + a z_1)^2 + (y_0 + b z_1)^2$. So there exists a rotation $\sigma$ around the $z$-axis that maps $(x_1, y_1, z_1)$ to the point $Q = (x_0 + a z_1, y_0 + b z_1, z_1)$ from $\ell$. So $(x_1, y_1, z_1)$ can be obtained by applying the inverse rotation $\sigma^{-1}$ to $\ell$. Note that $\sigma$ is unique, so only one of the rotated lines passes through $Q$.
Now we can solve the problem. In this case, since the hyperboloid has equation $3x^2 + 3y^2 - z^2 - 1 = 0 \iff x^2 + y^2 - \frac{1}{3}z^2 = \frac{1}{3}$ we can use $(x_0, y_0, 0) = (\frac{\sqrt{3}}{3}, 0, 0)$ and $(a, b, 1) = (0, \frac{\sqrt{3}}{3}, 1)$.

a. Since $(-a, -b, 1) = (0, -\frac{\sqrt{3}}{3}, 1)$, which is not parallel to $(a, b, 1)$, is also perpendicular to $(x_0, y_0, 0)$, the hyperboloid can be obtained rotating either the line $\ell_1: (\frac{\sqrt{3}}{3}, 0, 0) + t(0, \frac{\sqrt{3}}{3}, 1)$ or $\ell_2: (\frac{\sqrt{3}}{3}, 0, 0) + t(0, -\frac{\sqrt{3}}{3}, 1)$. So each point from $H$ belongs to at least two lines. We will prove that there are no other lines by showing that $H$ does not contain a line that is neither parallel to $(0, \frac{\sqrt{3}}{3}, 1)$ nor $(0, -\frac{\sqrt{3}}{3}, 1)$. Suppose it does. Since the sections of $H$ by horizontal planes are circles, this new line is not parallel to the $xy$-plane. So it contains a point in this plane, which we can suppose, without loss of generality, that is $(\frac{\sqrt{3}}{3}, 0, 0)$.
So suppose the line $r: (\frac{\sqrt{3}}{3}, 0, 0) + t(c, d, 1)$ is contained in $H$. So, for all $t \in \mathbb{R}$,
$$\begin{aligned} & 3 \left( \frac{\sqrt{3}}{3} + t c \right)^2 + 3 \cdot (t d)^2 - t^2 - 1 = 0 \\ \Leftrightarrow & (3c^2 + 3d^2 - 1)t^2 + \frac{2\sqrt{3}}{3} t c = 0 \\ \Leftrightarrow & \begin{cases} \frac{2\sqrt{3}}{3}c = 0 \\ 3c^2 + 3d^2 - 1 = 0 \end{cases} \quad \Leftrightarrow \quad \begin{cases} c = 0 \\ d = \pm \frac{\sqrt{3}}{3} \end{cases} \end{aligned}$$
So the support vector of $r$ should be $(0, \frac{\sqrt{3}}{3}, 1)$ or $(0, -\frac{\sqrt{3}}{3}, 1)$, and we are done.

b. We need to compute the angle that the vectors $(0, \pm\frac{\sqrt{3}}{3}, 1)$ define with the $xy$-plane: $\arctan \frac{1}{\sqrt{3}/3} = 60^\circ$.