(a) \( \frac{2^{x+2}-2^{x}}{2^{x}} \) (d) \( \frac{3^{m}+3^{m-1}}{2 \cdot 3^{m}} \) (g) \( \frac{2^{x}+2^{x+1}-2^{x+2}}{2^{x-3}} \)

**(a)** Rewrite the numerator by factoring \(2^x\): \[ 2^{x+2} - 2^x = 2^x \left(2^2 - 1\right) = 2^x (4 - 1) = 2^x \cdot 3. \] Then, the full expression becomes: \[ \frac{2^x \cdot 3}{2^x} = 3. \] **(d)** Rewrite the numerator by factoring \(3^{m-1}\): \[ 3^m + 3^{m-1} = 3^{m-1}\left(3 + 1\right) = 3^{m-1} \cdot 4. \] The expression is now: \[ \frac{4 \cdot 3^{m-1}}{2 \cdot 3^m}. \] Since \(3^m = 3^{m-1} \cdot 3\), cancel \(3^{m-1}\) in the numerator and denominator: \[ \frac{4}{2 \cdot 3} = \frac{4}{6} = \frac{2}{3}. \] **(g)** Factor the numerator by extracting \(2^x\): \[ 2^x + 2^{x+1} - 2^{x+2} = 2^x \left(1 + 2 - 4\right) = 2^x \cdot (-1) = -2^x. \] Thus, the expression becomes: \[ \frac{-2^x}{2^{x-3}}. \] Using the exponent rule \(\frac{2^x}{2^{x-3}} = 2^{x-(x-3)} = 2^3\), we have: \[ -2^3 = -8. \]