Find the equations of the tangent and normal to the curve y = x3 − 3x − 1 at the point (2, 3) and to the curve y = x2 − 2x − 1 when y = -1.

The equation of the tangent is $\boxed{y = 9x - 15}$.

Since the slope of the tangent is 0, the tangent line is horizontal. Therefore, the normal line is vertical.
A vertical line passing through $(1, -3)$ has the equation $x = 1$.
The equation of the normal is $\boxed{x = 1}$.

Substitute $y = -1$ into the curve's equation:

Factor out $x$:

This gives two x-coordinates: $x = 0$ or $x = 2$.
The points are $(0, -1)$ and $(2, -1)$.

At point $(0, -1)$:

Using $y - y_1 = m_t(x - x_1)$:

The equation of the tangent is $\boxed{y = -2x - 1}$.

The slope of the normal $m_n = -\frac{1}{m_t} = -\frac{1}{-2} = \frac{1}{2}$.
Using $y - y_1 = m_n(x - x_1)$:

The equation of the normal is $\boxed{y = \frac{1}{2}x - 1}$.

At point $(2, -1)$:

Using $y - y_1 = m_t(x - x_1)$:

The equation of the tangent is $\boxed{y = 2x - 5}$.

The slope of the normal $m_n = -\frac{1}{m_t} = -\frac{1}{2}$.
Using $y - y_1 = m_n(x - x_1)$:

The equation of the normal is $\boxed{y = -\frac{1}{2}x}$.

Set $\frac{dy}{dx} = 8$:

Substitute $x=8$ into the original equation:

The point is $(8, 36)$.

The gradient (slope of the tangent) at this point is $m_t = 8$.
The slope of the normal $m_n = -\frac{1}{m_t} = -\frac{1}{8}$.

Using the point-slope form $y - y_1 = m_n(x - x_1)$ with $(x_1, y_1) = (8, 36)$ and $m_n = -\frac{1}{8}$.

The equation of the normal is $\boxed{y = -\frac{1}{8}x + 37}$.