(i) The given equation is $k x^{2}+6 x+1=0$.

$\therefore D=6^{2}-4 \times k \times 1=36-4 k$

The given equation has real and distinct roots if $D>0$.

$\begin{array}{l}
\therefore 36-4 k>0 \\
\Rightarrow 4 k<36 \\
\Rightarrow k<9
\end{array}$

(ii) The given equation is $x^{2}-k x+9=0$.

$\therefore D=(-k)^{2}-4 \times 1 \times 9=k^{2}-36$

The given equation has real and distinct roots if $D>0$.

$\begin{array}{l}
\therefore k^{2}-36>0 \\
\Rightarrow(k-6)(k+6)>0 \\
\Rightarrow k<-6 \text { or } k>6
\end{array}$