Revision: a) 92 x 9^-2, b) 45 / 4^-2, c) 53 + 35, d) 50 - 14, e) (14.2)4 x (14.2)1/4, f) (14.2)4 + (14.2)1/4

Step 1: Solve part a). We use the exponent rule $a^m \times a^n = a^{m+n}$. $9^2 \times 9^{-2} = 9^{2 + (-2)}$ $= 9^{2-2}$ $= 9^0$ We use the exponent rule $a^0 = 1$. $= 1$ The answer for a) is $\boxed{1}$.

Step 2: Solve part b). We use the exponent rule $a^m \div a^n = a^{m-n}$. $4^5 \div 4^{-2} = 4^{5 - (-2)}$ $= 4^{5+2}$ $= 4^7$ Now we calculate the value of $4^7$. $4^7 = 4 \times 4 \times 4 \times 4 \times 4 \times 4 \times 4$ $= 16384$ The answer for b) is $\boxed{16384}$.

Step 3: Solve part c). We calculate each term separately and then add them. $5^3 = 5 \times 5 \times 5 = 125$ $3^5 = 3 \times 3 \times 3 \times 3 \times 3 = 243$ Now we add the results. $5^3 + 3^5 = 125 + 243$ $= 368$ The answer for c) is $\boxed{368}$.

Step 4: Solve part d). We use the exponent rule $a^0 = 1$. $5^0 = 1$ We calculate $1^4$. $1^4 = 1 \times 1 \times 1 \times 1 = 1$ Now we subtract the results. $5^0 - 1^4 = 1 - 1$ $= 0$ The answer for d) is $\boxed{0}$.

Step 5: Solve part e). We use the exponent rule $a^m \times a^n = a^{m+n}$. $(14.2)^4 \times (14.2)^{\frac{1}{4}} = (14.2)^{4 + \frac{1}{4}}$ To add the exponents, we find a common denominator. $4 + \frac{1}{4} = \frac{16}{4} + \frac{1}{4} = \frac{17}{4}$ So the expression simplifies to: $= (14.2)^{\frac{17}{4}}$ The answer for e) is $\boxed{(14.2)^{\frac{17}{4}}}$.

Step 6: Solve part f). We calculate each term separately and then add them. $(14.2)^4 = 14.2 \times 14.2 \times 14.2 \times 14.2 = 40603.0016$ $(14.2)^{\frac{1}{4}} = \sqrt[4]{14.2} \approx 1.940081$ Now we add the results. $(14.2)^4 + (14.2)^{\frac{1}{4}} \approx 40603.0016 + 1.940081$ $\approx 40604.941681$ The answer for f) is approximately $\boxed{40604.941681}$.