Hint: Use Ohm's law $I=\frac{V}{R_p}$ and the formula $\frac{1}{R_p}=\frac{1}{R}+\frac{1}{R_0}$ for parallel resistors to derive a straight line in a $I$-$\frac{1}{R}$ diagram
$$ I~=~V \left(\frac{1}{R}+\frac{1}{R_0}\right). $$
Here $R$ and $R_0$ are the variable and the fixed resistor, respectively. To answer OP's two questions:
The slope is $V=1{\rm Volt}$.
The intercept of the straight line on the vertical $I$-axis is $c=\frac{V}{R_0}$. The intercept of the straight line on the horizontal $\frac{1}{R}$-axis is $\frac{-1}{R_0}$.
A 242Ω resistor is in parallel with a 180Ω resistor, and a 240Ω
resistor is in series with the combination. A current of 22mA flows
through the 242Ω resistor. The current through the 180Ω resistor is
__mA.
Approaching this step by step, note that you can calculate the voltage across the $242 \Omega$ resistor since you're given the current through it. By Ohm's Law:
$$V_{R_{242}} = 22mA \cdot 242 \Omega$$
Now, you're also given that this resistor is in parallel with a $180\Omega$ resistor so the voltage across this resistor must be identical to the voltage across the $242 \Omega$ resistor.
Thus, and again by Ohm's Law, you can calculate the current through the $180\Omega$ resistor.
Two 24Ω resistors are in parallel, and a 43Ω resistor is in series
with the combination. When 78V is applied to the three resistors, the
voltage drop across the 24Ω resistor is___volts.
Again, approaching this step by step, the 2 parallel connected $24\Omega$ resistors can be combined into 1 equivalent resistor of resistance
$$R_{EQ} = \dfrac{24 \cdot 24}{24 + 24}$$
Now you can use voltage division to find the voltage across the equivalent resistance:
$$V_{R_{EQ}} = 78V \dfrac{R_{EQ}}{R_{EQ} + 43}$$
Can you take it from here?
Best Answer
What one refers to as "Ohm's law" is a linear relationship between the current and the potential difference, i.e., (see also this answer) $$ V = IR\text{ where } R=const $$ If $R$ is not a constant, the relationship is non-linear which is equivalent to saying that the Ohm's law does not hold.
One can (and does) introduce effective resistance/conductance that is dependent on the potential difference or current either as their ratio $$ R(V)=\frac{V}{I(V)} $$ or as a slope of the I-V characteristics $$ \frac{1}{R(V)}=\frac{dI(V)}{dV}. $$