Determine the total resistance of the circuit.

The circuit consists of a voltage source $18\,V$ connected to a series of resistors.
$R_1 = 4.7\,k\Omega$ is in series with the rest of the circuit.
$R_2 = 2.2\,k\Omega$ is in parallel with the series combination of $R_3$, $R_4$, $R_5$, and $R_6$.
Let's simplify the circuit from right to left.
$R_6 = 1.0\,k\Omega$ is connected to ground.
$R_5 = 1.0\,k\Omega$ is in series with $R_6$. So, $R_{56} = R_5 + R_6 = 1.0\,k\Omega + 1.0\,k\Omega = 2.0\,k\Omega$.
$R_4 = 2.2\,k\Omega$ is in parallel with $R_{56}$. So, $R_{456} = \frac{R_4 \times R_{56}}{R_4 + R_{56}} = \frac{2.2\,k\Omega \times 2.0\,k\Omega}{2.2\,k\Omega + 2.0\,k\Omega} = \frac{4.4\,k\Omega^2}{4.2\,k\Omega} \approx 1.0476\,k\Omega$.
$R_3 = 1.0\,k\Omega$ is in series with $R_{456}$. So, $R_{3456} = R_3 + R_{456} = 1.0\,k\Omega + 1.0476\,k\Omega = 2.0476\,k\Omega$.
$R_2 = 2.2\,k\Omega$ is in parallel with $R_{3456}$. So, $R_{23456} = \frac{R_2 \times R_{3456}}{R_2 + R_{3456}} = \frac{2.2\,k\Omega \times 2.0476\,k\Omega}{2.2\,k\Omega + 2.0476\,k\Omega} = \frac{4.50472\,k\Omega^2}{4.2476\,k\Omega} \approx 1.0605\,k\Omega$.
Finally, $R_1 = 4.7\,k\Omega$ is in series with $R_{23456}$.
Total Resistance $R_{total} = R_1 + R_{23456} = 4.7\,k\Omega + 1.0605\,k\Omega = 5.7605\,k\Omega$.

$$R_{total} \approx \boxed{\text{5.76 k}\Omega\text{}}$$

Point A is the voltage across the parallel combination of $R_2$ and $R_{3456}$.
The voltage drop across $R_1$ is $V_{R1} = I_{total} \times R_1 = 3.1247\,mA \times 4.7\,k\Omega = 14.686\,V$.

$$V_A = V_S - V_{R1} = 18\,V - 14.686\,V = 3.314\,V$$ $$V_A \approx \boxed{\text{3.31 V}}$$

The current flowing into the parallel combination of $R_2$ and $R_{3456}$ is $I_{total}$. This current splits.
The current through the branch containing $R_3, R_4, R_5, R_6$ is $I_{3456} = \frac{V_A}{R_{3456}} = \frac{3.314\,V}{2.0476\,k\Omega} \approx 0.0016185\,A = 1.6185\,mA$.
The voltage drop across $R_3$ is $V_{R3} = I_{3456} \times R_3 = 1.6185\,mA \times 1.0\,k\Omega = 1.6185\,V$.

$$V_B = V_A - V_{R3} = 3.314\,V - 1.6185\,V = 1.6955\,V$$ $$V_B \approx \boxed{\text{1.70 V}}$$

The current flowing into the parallel combination of $R_4$ and $R_{56}$ is $I_{3456}$. This current splits.
The voltage across $R_{456}$ is $V_B$.
The current through the branch containing $R_5$ and $R_6$ is $I_{56} = \frac{V_B}{R_{56}} = \frac{1.6955\,V}{2.0\,k\Omega} \approx 0.00084775\,A = 0.84775\,mA$.
The voltage drop across $R_5$ is $V_{R5} = I_{56} \times R_5 = 0.84775\,mA \times 1.0\,k\Omega = 0.84775\,V$.

$$V_C = V_B - V_{R5} = 1.6955\,V - 0.84775\,V = 0.84775\,V$$ $$V_C \approx \boxed{\text{0.85 V}}$$

The current $I_{3456}$ flows into the node before $R_4$ and $R_{56}$.
The voltage at point B is $V_B$. The voltage at point C is $V_C$.
The current through $R_4$ is $I_{R4} = \frac{V_B - V_{ground}}{R_4}$ if $R_4$ was connected to ground, but it's not.
The current through $R_4$ is the current that flows through $R_4$ itself.
The voltage across $R_4$ is $V_B - V_C$ if $R_4$ was between B and C, but it's not.
$R_4$ is in parallel with $R_5$ and $R_6$. The voltage across this parallel combination is $V_B$.
So, the current through $R_4$ is:

$$I_{R4} = \frac{V_B}{R_4} = \frac{1.6955\,V}{2.2\,k\Omega} \approx 0.00077068\,A = 0.77068\,mA$$ $$I_{R4} \approx \boxed{\text{0.77 mA}}$$