Electronic – What does 125VAC mean on a small indicator bulb

The fan requires 12V, 150mA (0.15A). About 1.8W as listed. That is for standard power. You can provide less volts, for less powerful blowing, at a relatively lower current draw.

As such you need a solar panel that can provide 12V and 0.15A, again 1.8W, in regular light. Unless you have 100% unobstructed direct sunlight, no clouds or anything, you will need a solar panel that can provide alot more than just 1.8W. So you need to over size it.

Since the bulb draws \$0.5\$A when \$1.5\$V is across it, its equivalent resistance by Ohm's Law is

$$R_{B} = \frac{1.5\text{V}}{0.5\text{A}} = 3\Omega$$

As you've stated in your question the solution is to use a series resistor \$R_{1}\$ and parallel (to the light bulb) resistor \$R_{2}\$. When the bulb is connected this creates a voltage divider composed of \$R_1\$ and \$R_2||R_B\$:

$$v_{s} = 1.5\text{V} = \frac{R_2 \parallel R_B}{R_2 \parallel R_B + R_1}6\text{V}$$

Also recall that

$$R_2 \parallel R_B = \frac{R_{2}R_{B}}{R_{2} + R_{B}}$$

When the bulb is disconnected the path through the bulb becomes an open circuit, which means that \$R_{B} \to \infty\$ (the actual bulb resistance is the same of course, but \$R_{B}\$ appears to be infinite to the circuit since the path is open). You have the same equation as above, except that \$v_{s} = 2\$V and \$R_2 \parallel R_B = R_2\$ (the equivalent resistance for a resistor in parallel with an infinite resistance is just the resistor value):

$$v_{s} = 2\text{V} = \frac{R_2}{R_2 + R_1}6\text{V}$$

You've got two equations (the two equations for \$v_{s}\$) and two unknowns (since \$R_{B}\$ is known) so you can solve the system for \$R_1\$ and \$R_2\$.