Substitute the given root into the equation to find k.

If $x = -2$ is a root of $x^2 + kx + 8 = 0$, then substituting $x=-2$ must satisfy the equation:

Substitute $k=6$ back into the original equation:

For a quadratic equation $ax^2 + bx + c = 0$, the sum of the roots is $-\frac{b}{a}$.
In our equation, $x^2 + 6x + 8 = 0$, we have $a=1$, $b=6$, $c=8$.
Let the two roots be $x_1$ and $x_2$. We are given $x_1 = -2$.
The sum of the roots is $x_1 + x_2 = -\frac{6}{1} = -6$.
Substitute $x_1 = -2$:

Alternatively, using the product of roots property:
The product of the roots is $x_1 \cdot x_2 = \frac{c}{a} = \frac{8}{1} = 8$.
Substitute $x_1 = -2$:

The other root is $\boxed{-4}$.

Let the first number be $x$.
The second number is greater than the first by 4, so the second number is $x+4$.

The product of the two numbers is 21.

The quadratic equation is $\boxed{x^2 + 4x - 21 = 0}$.

We need two numbers that multiply to -21 and add to 4. These numbers are 7 and -3.

Set each factor to zero:

Case 1: If $x = 3$.
The first number is 3.
The second number is $x+4 = 3+4 = 7$.
Check: $3 \times 7 = 21$.

Case 2: If $x = -7$.
The first number is -7.
The second number is $x+4 = -7+4 = -3$.
Check: $(-7) \times (-3) = 21$.

The two numbers are either $\boxed{\text{3 and 7}}$ or $\boxed{\text{-7 and -3}}$.