Determine the equivalent resistances between the points a and b for...

For network (a), let the nodes be $a$ (bottom-left), $b$ (bottom-right), C (top-left), and D (top-right).
The resistors between C and D, and D and $b$ are in series. Their equivalent resistance is $r + r = 2r$.
This combination is in parallel with the diagonal resistor between C and $b$. The equivalent resistance is $\dfrac{2r \times r}{2r + r} = \dfrac{2r}{3}$.
This is in series with the resistor between $a$ and C. The equivalent resistance of this upper branch is $r + \dfrac{2r}{3} = \dfrac{5r}{3}$.
Finally, this upper branch is in parallel with the resistor between $a$ and $b$. The total equivalent resistance is $\dfrac{\dfrac{5r}{3} \times r}{\dfrac{5r}{3} + r} = \dfrac{5r}{8}$.

For network (b), let the central node be O. The outer circle acts as a single conducting node, let's call it C.
Node $b$ is on the circle, so it is at the same potential as C.
Three resistors connect node O to the circle C. They are in parallel, giving an equivalent resistance of $\dfrac{r}{3}$.
The fourth resistor connects node $a$ to node O.
The total resistance between $a$ and $b$ is the series combination of the resistance from $a$ to O and from O to C.
$R_{ab} = r + \dfrac{r}{3} = \dfrac{4r}{3}$.

For network (c), the circuit is a balanced Wheatstone bridge.
The upper and lower arcs each consist of two resistors $r$ in series.
The vertical resistor $r$ connects the midpoints of these branches.
Due to symmetry, the bridge is balanced and no current flows through the central vertical resistor.
Removing the central resistor, the circuit consists of two parallel branches, each with resistance $r + r = 2r$.
The equivalent resistance is $\dfrac{2r \times 2r}{2r + 2r} = r$.

For network (d), the central node is $a$ and the outer circle is node $b$.
Four resistors are connected in parallel between node $a$ and node $b$.
The equivalent resistance is $\dfrac{r}{4}$.

For network (e), let the nodes along the main horizontal line be $N_1$ (which is $a$), $N_2$, $N_3$, and $N_4$ (which is $b$).
The circuit has resistors between $(N_1, N_2)$, $(N_2, N_3)$, $(N_3, N_4)$, $(N_1, N_3)$, and $(N_2, N_4)$, all of resistance $r$.
This forms a balanced Wheatstone bridge.
When a voltage is applied across $a$ and $b$, the potentials at $N_2$ and $N_3$ are equal due to symmetry.
No current flows through the resistor between $N_2$ and $N_3$, so it can be removed.
The circuit then consists of two parallel branches: one with resistors $(N_1, N_2)$ and $(N_2, N_4)$ in series, and another with resistors $(N_1, N_3)$ and $(N_3, N_4)$ in series.
Each branch has a resistance of $r + r = 2r$.
The equivalent resistance is $\dfrac{2r \times 2r}{2r + 2r} = r$.

The calculated equivalent resistances are $\dfrac{5}{8}r$, $\dfrac{4}{3}r$, $r$, $\dfrac{r}{4}$, and $r$ respectively.

Answer: (a) $\dfrac{5}{8} r$, (b) $\dfrac{4}{3} r$, (c) $r$, (d) $\dfrac{r}{4}$, (e) $r$