1. Problem 1: Given $q(p)=\frac{1}{1-p}$, find $q(-3)$. 2. We substitute $p=-3$ into the formula. 3. Compute $q(-3)=\frac{1}{1-(-3)}=\frac{1}{4}$. 4. Explanation: The minus sign cancels because $1-(-3)=1+3=4$, so the result is $\frac{1}{4}$. 5. Problem 2: Evaluate $2-2q^3+q^4-q$ when $q=2$. 6. Substitute $q=2$ into the expression. 7. Compute powers: $q^3=8$ and $q^4=16$. 8. Evaluate step by step: $2-2\cdot8+16-2=2-16+16-2=0$. 9. Therefore the value of the expression at $q=2$ is $0$. 10. Problem 3: Simplify $$(\frac{49}{16})^{-1/2}$$. 11. Use exponent rules: a negative exponent inverts the base and $1/2$ denotes the square root. 12. So $$(\frac{49}{16})^{-1/2}=(\frac{49}{16})^{-1\cdot\frac{1}{2}}=(\frac{16}{49})^{1/2}=\sqrt{\frac{16}{49}}=\frac{4}{7}$$. 13. Problem 4: Simplify the expression $$\frac{\log(m^p\sqrt{3})-n}{mn^2}$$ assuming the numerator is $\log(m^p\sqrt{3})-n$. 14. Use log properties: $\log(m^p\sqrt{3})=\log(m^p)+\log(\sqrt{3})=p\log m+\frac{1}{2}\log 3$. 15. Substitute into the numerator to get $p\log m+\frac{1}{2}\log 3-n$. 16. Therefore the whole expression simplifies to $$\frac{p\log m+\frac{1}{2}\log 3-n}{mn^2}$$. 17. Note: If a different grouping or formatting was intended for the original log expression, please clarify so I can adjust the simplification.