Answer
$$\frac{9}{4}$$
Work Step by Step
\begin{align*}
\lim _{h \rightarrow 0}\frac{\cos 3 h-1}{\cos 2h-1}&= \lim _{h \rightarrow 0} \frac{\cos 3 h-1}{h^2}\frac{h^2}{\cos 2h-1}\\
&= \lim _{h \rightarrow 0} \frac{\cos 3 h-1}{(3h)^2}\frac{(2h)^2}{\cos 2h-1}\\
&= \lim _{h \rightarrow 0} \frac{9}{4} \frac{1-\cos 3 h}{(3h)^2}\frac{(2h)^2}{1-\cos 2h}\\
&= \lim _{h \rightarrow 0}\frac{1-\cos 3 h}{(3h)^2}\frac{(2h)^2}{1-\cos 2h}\\
&= \frac{9}{4} \left(\frac{1}{2}\right)( 2).\\
&= \frac{9}{4}.
\end{align*}
Where we used the fact that $\lim _{h \rightarrow 0} \frac{1-\cos h}{h^2}=\frac{1}{2}$.
