Find the values of k and p given that line L1 (6x + 2y + 5k = 0) passes through (-3, -1) and is perpendicular to L2 (2px + 9y = 10). Also, find the equation of line L3 parallel to L2.

The equation of line $L_1$ is $6x + 2y + 5k = 0$.
Substitute $x = -3$ and $y = -1$ into the equation:

$$6(-3) + 2(-1) + 5k = 0$$ $$-18 - 2 + 5k = 0$$ $$-20 + 5k = 0$$ $$5k = 20$$ $$k = \frac{20}{5}$$ $$\boxed{k = 4}$$

Rearrange the equation $6x + 2y + 5k = 0$ into the form $y = mx + c$:

$$2y = -6x - 5k$$ $$y = -3x - \frac{5k}{2}$$

The gradient of $L_1$ is $m_1 = -3$.

The equation of line $L_2$ is $2px + 9y = 10$.
Rearrange this into the form $y = mx + c$:

$$9y = -2px + 10$$ $$y = -\frac{2p}{9}x + \frac{10}{9}$$

The gradient of $L_2$ is $m_2 = -\frac{2p}{9}$.

For perpendicular lines, the product of their gradients is $-1$: $m_1 \cdot m_2 = -1$.

$$(-3) \cdot \left(-\frac{2p}{9}\right) = -1$$ $$\frac{6p}{9} = -1$$ $$\frac{2p}{3} = -1$$ $$2p = -3$$ $$\boxed{p = -\frac{3}{2}}$$

The gradient of $L_2$ is $m_2 = -\frac{2p}{9}$.
Substitute $p = -\frac{3}{2}$:

$$m_2 = -\frac{2\left(-\frac{3}{2}\right)}{9} = -\frac{-3}{9} = \frac{3}{9} = \frac{1}{3}$$

Since line $L_3$ is parallel to $L_2$, their gradients are equal. So, $m_3 = \frac{1}{3}$.

The equation of $L_3$ can be written as $y = \frac{1}{3}x + C$, where $C$ is an arbitrary constant.
To convert this to the form $ax + by = c$, multiply by 3:

$$3y = x + 3C$$

Rearrange the terms:

$$x - 3y = -3C$$

Let $c = -3C$. Since $C$ is an arbitrary constant, $c$ is also an arbitrary constant.

$$\boxed{x - 3y = c \quad (\text{where } c \text{ is a constant})}$$

The gradient of $L_2$ is $m_2 = \frac{1}{3}$.
Let $\alpha$ be the acute angle the line makes with the x-axis.

$$\tan \alpha = m_2$$ $$\tan \alpha = \frac{1}{3}$$ $$\alpha = \arctan\left(\frac{1}{3}\right)$$ $$\alpha \approx 18.43^\circ$$

The obtuse angle $\theta$ is $180^\circ - \alpha$.

$$\theta = 180^\circ - 18.43^\circ$$ $$\boxed{\theta \approx 161.57^\circ}$$