Since 3 and $1 / 3$ are roots of the equation $a x^{2}-6 x+c=0$, we have:

$$
9 a-18+c=0 \Rightarrow 9 a+c=18 \text { and } \frac{a}{9}-2+c=0 \Rightarrow \frac{a}{9}+c=2
$$

Solving the system

$$
\left\{\begin{array}{l}
9 a+c=18 \\
\frac{a}{9}+c=2
\end{array} \quad \text { we obtain } \quad a=c=\frac{9}{5} . \text { Therefore, } a+c=\frac{18}{5}\right.
$$
2 - In a quadratic equation $a x^{2}+b x+c=0$, the sum of the roots is $-\frac{b}{a}$ and the product $\frac{c}{a}$. Therefore: $3+\frac{1}{3}=\frac{10}{3}=\frac{6}{a} \Rightarrow a=\frac{9}{5}$ and $3 \times \frac{1}{3}=1=\frac{c}{a} \Rightarrow a=c$. Thus, $a+c=2 \times \frac{9}{5}=\frac{18}{5}$.