To solve this problem, we need to analyze the circuit in two states: when the switch \( K \) is open and when it is closed.

### When \( K \) is Open:
Given:
- \( R_1 = 4 \Omega \)
- \( R_2 = 2 \Omega \)
- \( R_3 = 8 \Omega \)
- \( R_5 = 10 \Omega \)
- \( U = 12V \)
- \( I = 2A \) (current through the circuit when \( K \) is open)

When \( K \) is open, the circuit simplifies to a series circuit with \( R_1 \), \( R_2 \), and \( R_3 \).

The total resistance \( R_{total} \) when \( K \) is open:
\[ R_{total} = R_1 + R_2 + R_3 \]
\[ R_{total} = 4 \Omega + 2 \Omega + 8 \Omega = 14 \Omega \]

Using Ohm's law \( V = IR \):
\[ U = I \times R_{total} \]
\[ 12V = 2A \times 14 \Omega \]

This confirms that the given values are consistent.

### When \( K \) is Closed:
When \( K \) is closed, \( R_5 \) is added in parallel with \( R_3 \). We need to find the equivalent resistance of \( R_3 \) and \( R_5 \) in parallel.

The equivalent resistance \( R_{35} \) of \( R_3 \) and \( R_5 \) in parallel:
\[ \frac{1}{R_{35}} = \frac{1}{R_3} + \frac{1}{R_5} \]
\[ \frac{1}{R_{35}} = \frac{1}{8 \Omega} + \frac{1}{10 \Omega} \]
\[ \frac{1}{R_{35}} = \frac{10 + 8}{80} \]
\[ \frac{1}{R_{35}} = \frac{18}{80} \]
\[ R_{35} = \frac{80}{18} \]
\[ R_{35} \approx 4.44 \Omega \]

Now, the total resistance \( R_{total} \) when \( K \) is closed:
\[ R_{total} = R_1 + R_2 + R_{35} \]
\[ R_{total} = 4 \Omega + 2 \Omega + 4.44 \Omega \]
\[ R_{total} \approx 10.44 \Omega \]

The total current \( I \) through the circuit when \( K \) is closed:
\[ I = \frac{U}{R_{total}} \]
\[ I = \frac{12V}{10.44 \Omega} \]
\[ I \approx 1.15A \]

### Current through each resistor when \( K \) is closed:
1. **Current through \( R_1 \) and \( R_2 \)**:
Since \( R_1 \) and \( R_2 \) are in series with the total circuit, the current through them is the same as the total current:
\[ I_{R1} = I_{R2} = 1.15A \]

2. **Current through \( R_3 \) and \( R_5 \)**:
The voltage across \( R_3 \) and \( R_5 \) is the same because they are in parallel. This voltage is the same as the voltage drop across \( R_{35} \):
\[ V_{R35} = I \times R_{35} \]
\[ V_{R35} = 1.15A \times 4.44 \Omega \]
\[ V_{R35} \approx 5.11V \]

Using Ohm's law to find the current through \( R_3 \) and \( R_5 \):
\[ I_{R3} = \frac{V_{R35}}{R_3} \]
\[ I_{R3} = \frac{5.11V}{8 \Omega} \]
\[ I_{R3} \approx 0.64A \]

\[ I_{R5} = \frac{V_{R35}}{R_5} \]
\[ I_{R5} = \frac{5.11V}{10 \Omega} \]
\[ I_{R5} \approx 0.51A \]

### Summary:
- Current through \( R_1 \): \( I_{R1} \approx 1.15A \)
- Current through \( R_2 \): \( I_{R2} \approx 1.15A \)
- Current through \( R_3 \): \( I_{R3} \approx 0.64A \)
- Current through \( R_5 \): \( I_{R5} \approx 0.51A \)