Let the length of the part made of metal A be \( L_A \) and that of metal B be \( L_B \), with \( L_A + L_B = 20 \) cm.

Given:
- Expansion of metal A's rod = 0.075 cm, so expansion per cm for metal A = \( \frac{0.075}{20} = 0.00375 \) cm.
- Expansion of metal B's rod = 0.045 cm, so expansion per cm for metal B = \( \frac{0.045}{20} = 0.00225 \) cm.

The combined expansion of the third rod is 0.060 cm:

\[
L_A \cdot 0.00375 + L_B \cdot 0.00225 = 0.060
\]

Since \( L_B = 20 - L_A \), substitute:

\[
L_A \cdot 0.00375 + (20 - L_A) \cdot 0.00225 = 0.060
\]

Expanding and solving:

\[
0.00375L_A + 0.045 - 0.00225L_A = 0.060
\]
\[
0.0015L_A = 0.015
\]
\[
L_A = \frac{0.015}{0.0015} = 10 \text{ cm}
\]

So, the portion made of metal A has length 10 cm.

To find \( x \), we need to compare the Z scale with the Fahrenheit scale.

1. Range on the Z scale:
\[
65^\circ Z - (-14^\circ Z) = 65 + 14 = 79^\circ Z
\]

2. Range on the Fahrenheit scale:
The boiling and freezing points of water are 212ºF and 32ºF, respectively, so:
\[
212^\circ F - 32^\circ F = 180^\circ F
\]

3. Relationship between scales:
A change of \( 79^\circ Z \) corresponds to \( 180^\circ F \). Therefore:
\[
x = \frac{180}{79}
\]

So, \( x = \frac{180}{79} \).