1. Let $a$ and $b$ be the roots of the quadratic equation $x^{2}-x-1=0$. Then the value of $3 a^{2}+4 b+\frac{2}{a^{2}}$ is $\qquad$.
Given that $a b=-1, a+b=1$,
$$
a^{2}=a+1, b^{2}=b+1 \text {. }
$$

Therefore, $3 a^{3}+4 b+\frac{2}{a^{2}}=3 a^{3}+4 b+2 b^{2}$
$$
=3 a^{2}+3 a+6 b+2=6(a+b)+5=11
$$