$(a+b)^2-ab=a^2+b^2+ab\equiv 0\pmod{5}\Rightarrow 4a^2+4b^2+4ab\equiv 0\pmod{5}\Rightarrow (2a+b)^2+3b^2\equiv 0\pmod{5}$. We have $(2a+b)^2\equiv \{0,1,4\}\pmod{5}$ and $3b^2\equiv \{0,2,3\}\pmod{5}$. Then, we should have $(2a+b)^2\equiv 3b^2\equiv 0\pmod{5}$. Thus, $5|a,5|b$. We know that $a$ and $b$ are even, so $10|a$ and $10|b$. Hence, $a^3\equiv b^3\equiv 0\pmod{10}$. Then, I) is definitely incorrect. $M=a^2+b^2+ab\equiv 0+0+0=0\pmod{100}$. So II) is definitely correct.