Hello, struck!
1) P and Q are the points of intersection of the line
. . $\displaystyle \frac{x}{a} + \frac{y}{b} \:=\: 1,\quad a,b > 0$ . with the x and yaxes respectively.
The distance $\displaystyle PQ$ is 20, and the gradient of $\displaystyle PQ$ is 3.
Find the values of $\displaystyle a\text{ and }b. $ Did you make a sketch ? Code:

(0,b) * Q
\
 \
 \ 20
 \
 \ P
  +   *  
 (a,0)
Distance $\displaystyle PQ = 20.$
. . $\displaystyle \sqrt{(0a)^2 + (b0)^2} \:=\:20 \quad\Rightarrow\quad a^2 + b^2 \:=\:400\;\;{\color{blue}[1]}$
Slope of $\displaystyle PQ$ is 3.
. . $\displaystyle m \:=\:\frac{b0}{0a} \:=\:3\quad\Rightarrow\quad b \:=\:3a\;\;{\color{blue}[2]}$
Substitute [2] into [1]: .$\displaystyle a^2 + (3a)^2 \:=\:400 \quad\Rightarrow\quad 10a^2 \:=\:400$
. . $\displaystyle a^2 \:=\:40 \quad\Rightarrow\quad\boxed{ a \:=\:2\sqrt{10}}$
Substitute into [2]: .$\displaystyle b \:=\:3(2\sqrt{10}) \quad\Rightarrow\quad\boxed{ b \:=\:6\sqrt{10}}$