1. Find all real numbers a that satisfy \frac{1}{64a^3 + 7} - 7 = 0.
2. Find all values of a that satsify the equation \frac{a}{3} + 1 = \frac{a + 3}{a} + 1.
1. Starting with the given equation \frac{1}{64a^3 + 7} - 7 = 0, we can first simplify the left-hand side by adding 7 to both sides: \frac{1}{64a^3 + 7} = 7. Then we can invert both sides to get rid of the fraction: 64a^3 + 7 = \frac{1}{7}. Finally, we can solve for a by subtracting 7 from both sides and taking the cube root: a = \sqrt[3]{\frac{1}{7 \cdot 64}} - \frac{1}{4\sqrt[3]{7}}. Simplifying this expression gives a = -\frac{1}{4\sqrt[3]{7}}.
2. Starting with the given equation \frac{a}{3} + 1 = \frac{a + 3}{a} + 1, we can simplify the right-hand side to get \frac{a}{3} + 1 = \frac{a}{a} + \frac{3}{a} + 1. Combining like terms on the right-hand side gives \frac{a}{3} + 1 = 1 + \frac{3}{a}. Subtracting 1 from both sides gives \frac{a}{3} = \frac{3}{a}, and cross-multiplying gives a^2 = 9. Taking the square root of both sides gives a = \pm 3. Thus, the two solutions are a = 3 and a = -3.