[ThE-dArK-lOrD]

Idea for problem number $2$ of FE
From the problem, $f(f(x)+2g(x)+3f(y))=g(x)+2f(x)+3g(y)$
We get that $f(f(y)+2g(y)+3f(x))=g(y)+2f(y)+3g(x)$
If $f(x)+g(y) \ge g(x)+f(y) \rightarrow g(x)+2f(x)+3g(y)\ge g(y)+2f(y)+3g(x)$
Then $f(f(x)+2g(x)+3f(y)) \ge f(f(y)+2g(y)+3f(x))$
That is $f(x)+2g(x)+3f(y) \ge f(y)+2g(y)+3f(x)$ contradict with $f(x)+g(y) \ge g(x)+f(y)$

And for the case $f(x)+g(y) \le g(x)+f(y)$ , we can get contradict with similar way
So $f(x)+g(y) = g(x)+f(y)$ give us $f(x)=g(x)+k$
And the rest is easy.