To identify the two logic gates, we use the standard diode-resistor transistor logic behavior.

1. Gate (A)

In circuit (A), the arrangement is the usual diode logic OR gate:

• If either $V_A=5\,\text{V}$ or $V_B=5\,\text{V}$, the corresponding diode becomes forward biased.
• Then the output node rises to high voltage, so $V_0 \approx 5\,\text{V}$.
• Only when both inputs are $0\,\text{V}$ do both diodes remain off, and the output stays low.

Thus the truth table is:

This is an OR gate.

2. Gate (B)

In circuit (B), the transistor is used in the common-emitter switching mode:

• If input is high ($5\,\text{V}$), transistor turns ON.
• Then collector voltage falls close to $0\,\text{V}$, so output is low.
• If input is low ($0\,\text{V}$), transistor is OFF.
• Then the collector resistor pulls the output up to $5\,\text{V}$.

So the output is opposite to the input:

This is a NOT gate.

3. Final identification

• Circuit (A): OR gate
• Circuit (B): NOT gate

Therefore, the correct option is:

$\boxed{\text{B: OR and NOT Gate}}$

4. Comparison with stored correct answer

Stored correct answer: B

My derived answer: B

They match.