Find the equation of the tangent and normal to the circle x2 + 4x + y2 − 6y = 91 at the point (0, -7).

The given equation of the circle is $x^2 + 4x + y^2 - 6y = 91$.
We complete the square to find the standard form $(x-h)^2 + (y-k)^2 = r^2$.

$$(x^2 + 4x + 4) + (y^2 - 6y + 9) = 91 + 4 + 9$$ $$(x+2)^2 + (y-3)^2 = 104$$

The center of the circle is $C(-2, 3)$. The given point on the circle is $P(0, -7)$.

The formula for the gradient between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $m = \frac{y_2 - y_1}{x_2 - x_1}$.

$$m_{\text{radius}} = \frac{-7 - 3}{0 - (-2)} = \frac{-10}{2} = -5$$

The tangent line is perpendicular to the radius at the point of tangency. If two lines are perpendicular, the product of their gradients is $-1$.

$$m_{\text{tangent}} = -\frac{1}{m_{\text{radius}}} = -\frac{1}{-5} = \frac{1}{5}$$

Using the point-slope form $y - y_1 = m(x - x_1)$ with point $P(0, -7)$ and $m_{\text{tangent}} = \frac{1}{5}$:

$$y - (-7) = \frac{1}{5}(x - 0)$$ $$y + 7 = \frac{1}{5}x$$ $$y = \frac{1}{5}x - 7$$

The normal line is the line that passes through the center of the circle and the point of tangency. Therefore, its gradient is the same as the gradient of the radius.

$$m_{\text{normal}} = m_{\text{radius}} = -5$$

Using the point-slope form $y - y_1 = m(x - x_1)$ with point $P(0, -7)$ and $m_{\text{normal}} = -5$:

$$y - (-7) = -5(x - 0)$$ $$y + 7 = -5x$$ $$y = -5x - 7$$

The equation of the tangent is $\boxed{y = \frac{1}{5}x - 7}$.
The equation of the normal is $\boxed{y = -5x - 7}$.