Given a = 3+sqrt(5)2.

Given $a = \frac{3+\sqrt{5}}{2}$.
First, find $\frac{1}{a}$:

$$\frac{1}{a} = \frac{2}{3+\sqrt{5}}$$

Rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator, $3-\sqrt{5}$:

$$\frac{1}{a} = \frac{2}{3+\sqrt{5}} \times \frac{3-\sqrt{5}}{3-\sqrt{5}} = \frac{2(3-\sqrt{5})}{3^2 - (\sqrt{5})^2} = \frac{2(3-\sqrt{5})}{9-5} = \frac{2(3-\sqrt{5})}{4} = \frac{3-\sqrt{5}}{2}$$

Now, add $a$ and $\frac{1}{a}$:

$$a + \frac{1}{a} = \frac{3+\sqrt{5}}{2} + \frac{3-\sqrt{5}}{2} = \frac{3+\sqrt{5}+3-\sqrt{5}}{2} = \frac{6}{2} = 3$$

Substitute the value of $a + \frac{1}{a}$ found in Step 1:

$$a^2 + \frac{1}{a^2} = (3)^2 - 2 = 9 - 2 = 7$$

The value of $a^2 + \frac{1}{a^2}$ is $\boxed{\text{7}}$.

63. Simplify: $(256)^{(-4^{-2})}$

We know that $256 = 2^8$.

$$(2^8)^{-1/16} = 2^{8 \times (-\frac{1}{16})} = 2^{-\frac{8}{16}} = 2^{-\frac{1}{2}}$$

$$2^{-\frac{1}{2}} = \frac{1}{2^{1/2}} = \frac{1}{\sqrt{2}}$$

Rationalize the denominator:

$$\frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$

The simplified expression is $\boxed{\frac{\sqrt{2}}{2}}$.

64. Find the value of $\frac{4}{(216)^{-\frac{2}{3}}} + \frac{1}{(256)^{-\frac{3}{4}}} + \frac{2}{(243)^{-\frac{1}{5}}}$

The expression becomes:

$$4 \cdot (216)^{\frac{2}{3}} + 1 \cdot (256)^{\frac{3}{4}} + 2 \cdot (243)^{\frac{1}{5}}$$

For the first term, $4 \cdot (216)^{\frac{2}{3}}$:
Recognize that $216 = 6^3$.

$$4 \cdot (6^3)^{\frac{2}{3}} = 4 \cdot 6^{3 \times \frac{2}{3}} = 4 \cdot 6^2 = 4 \cdot 36 = 144$$

For the second term, $1 \cdot (256)^{\frac{3}{4}}$:
Recognize that $256 = 4^4$.

$$1 \cdot (4^4)^{\frac{3}{4}} = 1 \cdot 4^{4 \times \frac{3}{4}} = 1 \cdot 4^3 = 1 \cdot 64 = 64$$

For the third term, $2 \cdot (243)^{\frac{1}{5}}$:
Recognize that $243 = 3^5$.

$$2 \cdot (3^5)^{\frac{1}{5}} = 2 \cdot 3^{5 \times \frac{1}{5}} = 2 \cdot 3^1 = 2 \cdot 3 = 6$$

$$144 + 64 + 6 = 208 + 6 = 214$$

The value of the expression is $\boxed{\text{214}}$.