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\(a+b= \cfrac{√5+1}{√5-1}+\cfrac{√5-1}{√5+1}\)
\(= \cfrac{(√5+1)^2+(√5-1)^2}{(√5-1)(√5+1)}\)
\(= \cfrac{2(5+1)}{5-1}\)
\(= \cfrac{12}{4}=3\)

\(ab=\cfrac{√5+1}{√5-1}\times \cfrac{√5-1}{√5+1}=1\)

\(\cfrac{3a^2+5ab+3b^2}{3a^2-5ab+3b^2}\)
\(=\cfrac{3[a^2+b^2]+5ab}{3[a^2+b^2]-5ab}\)
\(=\cfrac{3[(a+b)^2-2ab]+5ab}{3[(a+b)^2-2ab]-5ab} \)
\(=\cfrac{3(a+b)^2-6ab+5ab}{3(a+b)^2-6ab-5ab} \)
\(=\cfrac{3(a+b)^2-ab}{3(a+b)^2-11ab} \)
\(=\cfrac{3\times(3)^2-1}{3\times(3)^2-11} \)
\(=\cfrac{27-1}{27-11} \)
\(=\cfrac{26}{16}=\cfrac{13}{8}=1\cfrac{5}{8} \) [Answer]