The first Problem, E:15 002, appeared in GMB_4/2016, at page 212 (the password for open file is ogeometrie ). I solved it myself in the post "Rezolvarea problemelor din Gazeta Matematica (seria B) Nr 4/2016 pentru clasa a VIII-a". The official solution to the problem appeared in GMB_10/2016, at pages 473-474; I have not yet compared it with my solution...
After almost 2000 problems, another version, by the same author, appears in GMB_6-7-8/2024, at page 369 (same pasword !!).
ANSWER CiP $a=2,\;\;b=2.$
Solution CiP
We add $2+2=4$ to both sides of the equation; so
$$\left ( \frac{2a-1}{b+2}+2 \right )+\left ( \frac{2b-1}{a+2}+2 \right )=\frac{6}{a+b}+4\;\Leftrightarrow$$
$$\Leftrightarrow\;\frac{2a+2b+3}{b+2}+\frac{2b+2a+3}{a+2}=\frac{6+4a+4b}{a+b};$$
since $2a+2b+3 \neq 0$, being an odd number, we can simplify the equation with him, and get
$$\frac{1}{b+2}+\frac{1}{a+2}=\frac{2}{a+b}\Leftrightarrow \frac{1}{b+2}-\frac{1}{a+b}=\frac{1}{a+b}-\frac{1}{a+2}\Leftrightarrow $$
$$\Leftrightarrow \frac{a-2}{(b+2)(a+b)}=\frac{2-b}{(a+b)(a+2)}\;\;\;\underset{a+b \neq 0}{\Leftrightarrow} \;\;\;\frac{a-2}{b+2}=\frac{2-b}{a+2}\Leftrightarrow$$
$$\Leftrightarrow a^2-4=4-b^2\Leftrightarrow a^2+b^2=8;$$
the last equation is possible with integer numbers only if $a^2=4,\;b^2=4$ and, because $a,b \neq -2$, we have $a=2$ and $b=2.$
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