Solve the following exponential equations: 2^(2x+4)=1, 2^(4x+2)=64, 2^(x+3)=4^x, 3^(2x+7)=81.

$$2^{2x+4} = 1$$

Since $1 = 2^0$, we have:

$$2^{2x+4} = 2^0$$

$$2x = -4$$ $$x = \frac{-4}{2}$$ $$x = -2$$

The student's solution for this problem is correct.

$$\boxed{x = -2}$$

:

$$2^{4x+2} = 64$$

Since $64 = 2^6$, we have:

$$2^{4x+2} = 2^6$$

$$4x = 6-2$$ $$4x = 4$$ $$x = \frac{4}{4}$$ $$x = 1$$

The student's solution for this problem is correct.

$$\boxed{x = 1}$$

:

$$2^{x+3} = 4^x$$

Since $4 = 2^2$, we have:

$$2^{x+3} = (2^2)^x$$

Using the exponent rule $(a^m)^n = a^{mn}$:

$$2^{x+3} = 2^{2x}$$

$$3 = 2x - x$$ $$3 = x$$ $$\boxed{x = 3}$$

The student's solution for this problem is incorrect.

:

$$3^{2x+7} = 81$$

Since $81 = 3^4$, we have:

$$3^{2x+7} = 3^4$$

$$2x = 4-7$$ $$2x = -3$$ $$x = \frac{-3}{2}$$ $$x = -1.5$$

The student's solution for this problem is correct.

$$\boxed{x = -1.5}$$