To find the equation of the tangent and normal to the curve x² + y² = 11 at the point (3, 1).

The equation of the curve is $x^2 + y^2 = 11$.
Differentiate both sides with respect to $x$:

$$\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(11)$$ $$2x + 2y \frac{dy}{dx} = 0$$ $$2y \frac{dy}{dx} = -2x$$ $$\frac{dy}{dx} = -\frac{2x}{2y}$$ $$\frac{dy}{dx} = -\frac{x}{y}$$

Substitute $x=3$ and $y=1$ into the derivative:

$$m_{\text{tangent}} = \frac{dy}{dx} \Big|_{(3,1)} = -\frac{3}{1} = -3$$

The equation of a line is $y - y_1 = m(x - x_1)$.
Using the point $(x_1, y_1) = (3, 1)$ and slope $m_{\text{tangent}} = -3$:

$$y - 1 = -3(x - 3)$$ $$y - 1 = -3x + 9$$ $$y = -3x + 10$$

Rearrange into the standard form $Ax + By + C = 0$:

$$3x + y - 10 = 0$$

The normal line is perpendicular to the tangent line. The product of their slopes is $-1$.

$$m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}}$$ $$m_{\text{normal}} = -\frac{1}{-3} = \frac{1}{3}$$

Using the point $(x_1, y_1) = (3, 1)$ and slope $m_{\text{normal}} = \frac{1}{3}$:

$$y - 1 = \frac{1}{3}(x - 3)$$

Multiply by 3 to clear the fraction:

$$3(y - 1) = x - 3$$ $$3y - 3 = x - 3$$ $$3y = x$$

Rearrange into the standard form $Ax + By + C = 0$:

$$x - 3y = 0$$

The equation of the tangent line is $\boxed{3x + y - 10 = 0}$.
The equation of the normal line is $\boxed{x - 3y = 0}$.