A) The lines will be parallel to each other if their slopes/gradients are equal, hence we get
$$\frac{5}{6}=\frac{2}{p}\iff p=\frac{12}{5}$$
B) The lines will be perpendicular to each other if the product of their slopes/gradients is $-1$, hence we get
$$\frac{5}{6}\times \frac{2}{p}=-1\iff p=\frac{-5}{3}$$
Notice that in two dimensions, the slope (when defined) is the ratio of the components of the directional vector:
$$ \langle x,y\rangle = \langle x_0, y_0\rangle + t\langle m_x, m_y\rangle \land m_x \neq 0
\implies y = \frac{m_y}{m_x} x + \left(y_0 - \frac{m_y}{m_x}x_0 \right). $$
Note that flipping the signs of $m_x$ and $m_y$ simultaneously does nothing to the line at all.
The line given by
$\langle x,y\rangle = \langle x_0, y_0\rangle + t\langle -m_x, -m_y\rangle$
is exactly the same as the line given by
$\langle x,y\rangle = \langle x_0, y_0\rangle + t\langle m_x, m_y\rangle.$
The "negative reciprocal" rule is actually telling you to change
$\frac{m_y}{m_x}$ to $-\frac{m_x}{m_y}.$ A simple way to do this is to swap the two components and negate just one of them:
$\langle x,y\rangle = \langle x_0, y_0\rangle + t\langle m_y, -m_x\rangle.$
Notice how the dot product shows that the new vector $\langle m_y, -m_x\rangle$
is perpendicular to the old vector $\langle m_x, m_y\rangle$:
$$
\langle m_x, m_y\rangle \cdot \langle m_y, -m_x\rangle = m_x m_y + m_y(-m_x) = 0.
$$
In short, even in two dimensions, the rule is not to take the negative reciprocal of each component of the direction vector simultaneously.
In two dimensions there is only one line perpendicular to a given line,
but in three dimensions, there are many lines perpendicular to a given line.
We therefore have more choices for the direction vector for the perpendicular line.
Given a direction vector $\langle m_x, m_y, m_z\rangle,$
where $m_x$ and $m_y$ are not both zero,
one possible choice of a perpendicular vector is
$\langle m_y, -m_x, 0\rangle,$
which is an application of the negative reciprocal rule to the first two dimensions;
this still works, even in three dimensions, as long as we make sure not to allow the third dimension to make the dot product non-zero.
(That is why the third component of the perpendicular vector is zero.)
Another vector perpendicular to both of these is
$\langle m_x m_z, m_y m_z, -(m_x^2 + m_y^2)\rangle.$
Best Answer
Your answer to the first question is correct.
For the second question, your error, as pointed out in the comment above, was that you accidentally cancelled the $b$'s, even though they had opposite signs. You have two equations:
$$8 = 6m+b \tag{1}$$
$$8 = -\left(\frac{1}{m}\right)6-b \tag{2}$$
There are multiple ways this system can be solved. For instance, you can isolate $b$ in equation $(1)$ and plug it in equation $(2)$.
$$8 = 6m+b \iff \color{blue}{b = 8-6m} \tag{1}$$
$$\implies 8 = -\left(\frac{1}{m}\right)6-\color{blue}{(8-6m)}$$
Can you take it on from here?