Step 1: Recall the two telescope facts.
In normal adjustment the magnifying power is $M = \frac{f_o}{f_e}$ and the tube length is $L = f_o + f_e$, where $f_o$ is the objective focal length and $f_e$ the eyepiece focal length.
Step 2: List what we know.
$M = 5$ and $L = 36$ cm.
Step 3: Use the magnification to link the two lenses.
From $M = 5$ we get $f_o = 5 f_e$.
Step 4: Substitute into the length equation.
\[ f_o + f_e = 36 \;\Rightarrow\; 5f_e + f_e = 36 \]
Step 5: Solve for the eyepiece.
$6 f_e = 36$, so $f_e = 6$ cm.
Step 6: Back-substitute for the objective.
$f_o = 5 \times 6 = 30$ cm. The objective is the longer focal length lens, as expected for a telescope.
\[ \boxed{f_o = 30\ \text{cm},\quad f_e = 6\ \text{cm}} \]