The intersection of the two lines is not $(13,4)$. Plugging that point into the first line gives $8$ and into the second gives $4$ for the left hand side, not $0$, so you need to redo the solution for the point. The $x$ coordinate is correct, but not $y$. Then there are many lines through that point and you are to use the distance information to pick two of them. If the intersection point is $(a,b)$, the point slope formula would be $y-b=m(x-a)$ and you need to use your formula for the distance between $(1,4)$ and this line to pick $m$.
Added: I meant the line through $(13,8)$ of slope $m$, so $y-8=m(x-13)$ or $-mx+y+13m-8=0$. You need the distance from $(1,4)$ to this line to be $4$. Mathworld gives the distance from the point $(x_0,y_0)$ to the line $ax+by+c=0$ as $\frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}$ so you have $4=\frac{|-m+4+13m-8|}{\sqrt{m^2+1}}$. Moving the $\sqrt{m^2+1}$ to the other side and squaring, you should find two values for $m$.
What we have is a line with positive slope, passing through the point $(-2, 5)$, with the length of the segment on this line from the point where it intersects the negative $x$-axis at $(x_0, 0)$ and the positive $y$-axis at $(0, y_0)$ equal to $7\sqrt 2$.
We know then, that $$x_0^2 + y_0^2 = (7\sqrt 2)^2 = 98\tag{1}$$
We also know that the distance from $(x_0, 0)$ to $(-2, 5)$ plus the distance from $(-2, 5)$ to the point $(0, y_0)$ is equal to $7\sqrt 2$: $$\sqrt {25 + (x_0+2)^2} + \sqrt{4+(y_0-5)^2} = 7\sqrt 2\tag{2}.$$
Solving $(1)$ and $(2)$ simultaneously gives us the real solutions $x_0 = -7$, and $y_0 = 7$. With these values, we must have that the slope of each of these "segments" comprising the entire length are equal: $$m = \frac{5-0}{-2 - x_0} = \frac{5-y_0}{-2-0}= 1$$
Now, given your point on the line $(-2, 5)$ and the slope of $m = 1$, we can construct the equation of the line: $$y - 5 = x + 2 \iff y = x+7$$
Best Answer
Equation of lines that passes through $(8,6)$ is $y=ax+6-8a$ and with axes that has common points $(0,6-8a),(8-\frac{6}{a},0)$. Then the surface of right triangle is $$S=\frac{(8-\frac{6}{a})(6-8a)}{2}=12$$
after simplification we get equation
$$16a^2-30a+9=0$$ with solutions $$a_1=6,a_2=24$$ there exists two lines with such proppertie $$l_1:y=6x-42$$ $$l_2:y=24x-186$$