To solve the given circuit, we need to find the equivalent resistance \( R_{AB} \) and the currents \( i_1 \), \( i_2 \), and \( i_3 \).

### a) Tính \( R_{AB} \)?

The circuit consists of \( R_1 \) in series with the parallel combination of \( R_2 \) and \( R_3 \).

1. **Calculate the equivalent resistance of \( R_2 \) and \( R_3 \) in parallel:**

\[ \frac{1}{R_{23}} = \frac{1}{R_2} + \frac{1}{R_3} \]

Given \( R_2 = 10 \Omega \) and \( R_3 = 10 \Omega \):

\[ \frac{1}{R_{23}} = \frac{1}{10} + \frac{1}{10} = \frac{2}{10} = \frac{1}{5} \]

\[ R_{23} = 5 \Omega \]

2. **Calculate the total resistance \( R_{AB} \):**

\[ R_{AB} = R_1 + R_{23} \]

Given \( R_1 = 15 \Omega \):

\[ R_{AB} = 15 \Omega + 5 \Omega = 20 \Omega \]

So, \( R_{AB} = 20 \Omega \).

### b) \( i_1 \), \( i_2 \), \( i_3 \)?

1. **Calculate the total current \( i_1 \) using Ohm's Law:**

\[ i_1 = \frac{U_{AB}}{R_{AB}} \]

Given \( U_{AB} = 35V \) and \( R_{AB} = 20 \Omega \):

\[ i_1 = \frac{35V}{20 \Omega} = 1.75A \]

2. **Calculate the voltage across \( R_2 \) and \( R_3 \):**

Since \( R_2 \) and \( R_3 \) are in parallel, the voltage across them is the same and equal to the voltage drop across \( R_{23} \):

\[ V_{R_{23}} = i_1 \times R_{23} \]

\[ V_{R_{23}} = 1.75A \times 5 \Omega = 8.75V \]

3. **Calculate the currents \( i_2 \) and \( i_3 \):**

Using Ohm's Law for each resistor:

\[ i_2 = \frac{V_{R_{23}}}{R_2} = \frac{8.75V}{10 \Omega} = 0.875A \]

\[ i_3 = \frac{V_{R_{23}}}{R_3} = \frac{8.75V}{10 \Omega} = 0.875A \]

### Summary:
- \( R_{AB} = 20 \Omega \)
- \( i_1 = 1.75A \)
- \( i_2 = 0.875A \)
- \( i_3 = 0.875A \)