Rewrite 0.25 and 16 with a common base.

$$0.25 = \frac{1}{4} = 4^{-1}$$ $$16 = 4^2$$

$$4^{-(x+1)} = 4^2$$ $$4^{-x-1} = 4^2$$

$$-x = 2 + 1$$ $$-x = 3$$ $$x = -3$$

The solution is $\boxed{-3}$.

2. Solve for $x$: $3(3)^x = 27$

$$3^x = \frac{27}{3}$$ $$3^x = 9$$

$$x = 2$$

The solution is $\boxed{2}$.

4. Solve for $x$: $10^{-x} = 0.000001$

$$x = 6$$

The solution is $\boxed{6}$.

5. Solve: $(3^x)^2 + 2(3^x) - 3 = 0$

Substitute $y$ into the equation.

$$y^2 + 2y - 3 = 0$$

$$y+3 = 0 \implies y = -3$$ $$y-1 = 0 \implies y = 1$$

Case 1: $3^x = -3$
An exponential function with a positive base (3) can never be negative. Thus, there is no real solution for $x$ in this case.

Case 2: $3^x = 1$

$$3^x = 3^0$$

Equate the exponents.

$$x = 0$$

The solution is $\boxed{0}$.

6. Solve: $5^{2x+1} - 26(5^x) + 5 = 0$

Substitute $y$ into the equation.

$$5y^2 - 26y + 5 = 0$$

$$5y^2 - 25y - y + 5 = 0$$ $$5y(y - 5) - 1(y - 5) = 0$$ $$(5y - 1)(y - 5) = 0$$

$$5y - 1 = 0 \implies 5y = 1 \implies y = \frac{1}{5}$$ $$y - 5 = 0 \implies y = 5$$

Case 1: $5^x = \frac{1}{5}$

$$5^x = 5^{-1}$$

Equate the exponents.

$$x = -1$$

Case 2: $5^x = 5$

$$5^x = 5^1$$

Equate the exponents.

$$x = 1$$

The solutions are $\boxed{-1, 1}$.