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Step 1: Identify the formula for equivalent resistance in series.

$R_{eq} = R_1 + R_2 + R_3$

Substitute $R_1 = 2\,\Omega$, $R_2 = 3\,\Omega$, $R_3 = 5\,\Omega$:

$R_{eq} = 2\,\Omega + 3\,\Omega + 5\,\Omega = 10\,\Omega$

Step 2: Calculate total current using Ohm's law.

$I = \frac{V}{R_{eq}}$

Substitute $V = 20\,\mathrm{V}$, $R_{eq} = 10\,\Omega$:

$I = \frac{20\,V}{10\,\Omega} = 2\,\mathrm{A}$

Final answer for 1.1: $2\,\mathrm{A}$ $\boxed{2\,A}$

Step 1 (1.2): Current is same in series circuit, so $I = 2\,\mathrm{A}$ in all resistors.

Voltage across $R_1$:

$V_1 = I R_1 = 2\,A \times 2\,\Omega = 4\,\mathrm{V}$

Final answer for 1.2: $4\,\mathrm{V}$ $\boxed{4\,V}$

Step 1 (1.3): Power in $R_2$.

$P = I^2 R_2$

Substitute $I = 2\,\mathrm{A}$, $R_2 = 3\,\Omega$:

$P = (2\,A)^2 \times 3\,\Omega = 12\,\mathrm{W}$

Final answer for 1.3: $12\,\mathrm{W}$ $\boxed{12\,W}$

Step 1 (2.1): Equivalent resistance for parallel.

$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$

Substitute $R_1 = 4\,\Omega$, $R_2 = 6\,\Omega$:

$\frac{1}{R_{eq}} = \frac{1}{4\,\Omega} + \frac{1}{6\,\Omega} = \frac{3}{12\,\Omega} + \frac{2}{12\,\Omega} = \frac{5}{12\,\Omega}$

$R_{eq} = \frac{12}{5}\,\Omega = 2.4\,\Omega$

Final answer for 2.1: $2.4\,\Omega$ $\boxed{2.4\,\Omega}$

Step 1 (2.2): Total current.

$I_{total} = \frac{V}{R_{eq}} = \frac{12\,V}{2.4\,\Omega} = 5\,\mathrm{A}$

Final answer for 2.2: $5\,\mathrm{A}$ $\boxed{5\,A}$

Step 1 (2.3): Current through $R_1$.

$I_1 = \frac{V}{R_1} = \frac{12\,V}{4\,\Omega} = 3\,\mathrm{A}$

Final answer for 2.3: $3\,\mathrm{A}$ $\boxed{3\,A}$