${\text{D}}_f=⟨-4,\to ⟩$, ${\text{B}}_{\left(f\right)}=ℝ$, vertical asymptote $x=-4$.

$f\left(x\right)=0$ gives you $log(x+4)=\frac{1}{3}$ and therefore $x+4={10}^{\frac{1}{3}}$, so $x=\sqrt[3]{10}-4$.
Graph: $x>\sqrt[3]{10}-4$.

${\text{D}}_g=⟨1,\to ⟩$, ${\text{B}}_g=ℝ$, vertical asymptote $x=1$.

$g\left(x\right)=-14$ gives you ${}^{\frac{1}{3}}log(x-1)=-2$ and therefore $x-1={\left(\frac{1}{3}\right)}^{-2}=9$ so $x=10$.
Graph: $1<x\le 10$.

${}^{3}log\left(x\right)={}^{3}log\left(5^2\right)$, so $x=5^2=25$.

${}^{\frac{1}{3}}log\left(x\right)={}^{\frac{1}{3}}log(5\cdot 2)$, so $x=10$.

${}^{2}log\left(x\right)=5$, so $x=2^5=32$.

${}^{5}log\left(x\right)={}^{5}log\left(5^3\right)+{}^{5}log\left(3^4\right)={}^{5}log(5^3\cdot 3^4)$, so $x=125\cdot 81=10125$.

$x=5(2-x)$ gives you $x=\frac{10}{6}=\frac{5}{3}$.

${}^{5}log\left(x\right)={}^{5}log\left(5^3\right)+{}^{5}log\left(x^4\right)={}^{5}log(5^3\cdot x^4)$, therefore $x=125x^4$, and so $x(125x^3-1)=0$ such that $x=0\vee x=0.2$.
Since $x=0$ is not allowed, the answer is $x=0.2$.

${\text{D}}_f=⟨0,\to ⟩$, ${\text{B}}_f=ℝ$, vertical asymptote $x=0$.
${\text{D}}_g=⟨\leftarrow ,4⟩$, ${\text{B}}_g=ℝ$, vertical asymptote $x=4$.

$log\left(x\right)=-1+log(4-x)$ gives you $log\left(x\right)=log\left(0.1(4-x)\right)$.
This means: $x=0.4-0.1x$ and therefore $x=\frac{0.4}{1.1}=\frac{4}{11}$.

$0<x\le \frac{4}{11}$

$\frac{4}{11}<x<4$

$q=5-3^{15-p}$

$q=200\cdot {10}^{\frac{(p-600)}{15}}$