Let's solve each of the exponential equations step by step, without using logarithms.

1. $4^{x+3} = 1$

For any base $a$, $a^0 = 1$. Thus:

$4^{x+3} = 4^0 \implies x + 3 = 0$

Solving for $x$:

$x = -3$

2. $5^{3-2x} = 125$

First, express $125$ as a power of $5$:

$125 = 5^3$

Now, equate the exponents:

$3 - 2x = 3$

$-2x = 0 \implies x = 0$

3. $3^{1-2x} = 243$

First, express $243$ as a power of $3$:

$243 = 3^5$

Now, equate the exponents:

$1 - 2x = 5$

$-2x = 4 \implies x = -2$

4. $3^{2x} = 3^{-x}$

Since the bases are the same, set the exponents equal to each other:

$2x = -x$

$3x = 0 \implies x = 0$

5. $8^{2x+5} = 4096$

Express $4096$ as a power of $8$:

$8^4 = 4096$

Now, equate the exponents:

$2x + 5 = 4$

$2x = -1 \implies x = -\frac{1}{2}$

6. $4^x = 64$

Express $64$ as a power of $4$:

$64 = 4^{3/2}$

Now, equate the exponents:

[ x = \frac 3