1. For set $A$, the condition $(x + 3)(x - 6) \geq 0$ holds if $x \leq -3$ or $x \geq 6$. Hence, $A = \{ x \mid x \leq -3 \text{ or } x \geq 6 \}$.

For set $B$, the inequality $\frac{x + 2}{x - 14} < 0$ implies that the numerator and denominator have opposite signs, which holds if $-2 < x < 14$. Therefore, $B = \{ x \mid -2 < x < 14 \}$.

The complement of $B$ in the real numbers, $\complement_R B$, includes all the real numbers not in $B$, which are $x \leq -2$ or $x \geq 14$. So, $\complement_R B = \{ x \mid x \leq -2 \text{ or } x \geq 14 \}$.

The intersection $A \cap \complement_R B$ then consists of all elements that are both in $A$ and in $\complement_R B$, which gives us $\boxed{ \{ x \mid x \leq -3 \text{ or } x \geq 14 \} }$.

2. For set $E$, given $E = \{ x \mid 2a < x < a + 1 \}$ and $E \subseteq B$, we need to consider two scenarios:

Case I: $E$ is an empty set. This happens when the interval defined by $E$ makes no sense, i.e., when the lower bound is greater than or equal to the upper bound. This can be expressed as $2a \geq a + 1$, which simplifies to $a \geq 1$.

Case II: $E$ is not empty. For $E$ to be a subset of $B$, the interval $2a < x < a + 1$ must lie within the interval defined by $B$ which is $-2 < x < 14$. Therefore, the following inequalities must hold:
$$\begin{cases} 2a \geq -2 \\ a + 1 \leq 14 \end{cases}$$
Solving these inequalities gives: $a \geq -1$ and $a < 13$.

Combining the results from both cases, the range for $a$ is:
$$\boxed{ a \geq -1 }$$