You only did 11-16 I am expecting 17-20

The alternating current is $I = 3 \sin (50t)$. From this, the peak current is $I_{peak} = 3 \text{ A}$.
The resistive load is $R = 10 \text{ Ω}$.
The mean power dissipated in a resistive load is $P_{mean} = I_{rms}^2 R$.

The relationship between RMS current ($I_{rms}$) and peak current ($I_{peak}$) is $I_{rms} = \frac{I_{peak}}{\sqrt{2}}$.

The mean power dissipated is $\boxed{\text{}45 \text{ W}}$.

Power of the heater $P = 4.0 \text{ kW} = 4000 \text{ W}$.
RMS voltage of the supply $V_{rms} = 240 \text{ V}$.
The power in an AC circuit is $P = V_{rms} I_{rms}$.
The relationship between RMS current and peak current is $I_{peak} = I_{rms} \sqrt{2}$.

The peak current drawn is $\boxed{\text{}23.6 \text{ A}}$.

Primary voltage $V_p = 220 \text{ V}$.
Secondary voltage $V_s = 7.5 \text{ V}$.
Number of turns on the secondary coil $N_s = 140$ turns.
For an ideal transformer, the ratio of voltages is equal to the ratio of turns: $\frac{V_p}{V_s} = \frac{N_p}{N_s}$.

Since the number of turns must be an integer, we round to the nearest whole number.
The number of turns on the primary coil is $\boxed{\text{}4107 \text{ turns}}$.

Primary voltage $V_p = 240 \text{ V}$.
Secondary voltage $V_s = 15 \text{ V}$.
Secondary current $I_s = 2.0 \text{ A}$.
For an ideal transformer, the power in the primary coil equals the power in the secondary coil: $P_p = P_s$.
This can be written as $V_p I_p = V_s I_s$.

The current drawn from the primary coil is $\boxed{\text{}0.125 \text{ A}}$.