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(a) let's write an equation to solve this problem. 

Let's let w be the total number of watermelons and m be the total number of mangoes. 

Using the problem, we have the system of equations

$w=m+13\\ 2/5w = 4/9m$

Subsituting out w with m from the first equation, we get

$\frac{2}{5}(m+13) = \frac{4}{9}m$

Solving this equation yields 

$m=117$. Since there are 13 more watermelons, then $w=130$

Next, let's find how many of each there were. Since $2/5(130) = 52$, we have

$52*2=104$

So 104 is our final answer. 

Thanks! :)

(b) We know there were 52 watermelons and 52 mangoes remaining. 

There are also 130 watermelones and 117 mangoes. 

Thus, we just have

$130-52=78\\ 117-52=65$

Adding the two numbers up, we find that $78+65=143$

This means they sold a total of 143 watermelons and mangoes. 

Since each cost 3.60, we have

$3.60 *143 = 514.8$

Thus, they collected 514.8 dollars. 

Thanks! :)