To solve for \( R_3 \) and \( R_5 \), given the specifications for the bulbs and assuming they operate under normal conditions, you can follow these steps:

1. **Find the resistance of each bulb:**
- For bulb \( Đ1 \) (3V, 1.5W):
\[
I_1 = \frac{P}{V} = \frac{1.5}{3} = 0.5 \text{ A}
\]
\[
R_1 = \frac{V}{I} = \frac{3}{0.5} = 6 \, \text{Ohm}
\]

- For bulb \( Đ2 \) (6V, 3W):
\[
I_2 = \frac{P}{V} = \frac{3}{6} = 0.5 \text{ A}
\]
\[
R_2 = \frac{V}{I} = \frac{6}{0.5} = 12 \, \text{Ohm}
\]

2. **Replace bulbs with their resistances in the circuit:**
- \( R_{D1} = 6 \, \Omega \)
- \( R_{D2} = 12 \, \Omega \)

3. **Analyze the circuit:**
- Since \( D1 \) and \( D2 \) are in parallel, and the combination is in series with \( R_3 \) and \( R_4 \), find the total current using:

The total voltage across the series is \( U_{ab} = 15V \). Calculating the total resistance \( R_T \) in the series path (ignoring \( R_3 \) for now):
\[
R_T = R_4 + R_5
\]

4. **Solve for \( R_5 \) assuming full voltage at bulbs:**

From Parallel arrangements:
\[
V_{parallel} = 6V \text{ (as it powers both bulbs)}
\]

The remaining voltage across \( R_3 + R_4 \) (i.e., \(15V - 6V = 9V\)).

5. **Solving for R3:**

Using Ohm’s Law:
\[
V = I \times R
\]

\[
9 = I \times (R_3 + R_4)
\]

Where current \( I \) is the same as bulb current:
\[
0.5A = \frac{9}{(R_3 + R_4)}
\]
Solve for \( R_3 \).

6. **Solve for R5:**

Using the knowledge that the total parallel voltage must equal the power requisite of bulbs:
\[
R_5 = \frac{6V}{0.5A} - \text{because it shares with D1}
\]

Verify all calculations align with given circuit specs to assure \( R_3 \), and \( R_5 \) values maintain desired normal bulb operation.