We use the spontaneity criterion:

$$\Delta G = \Delta H - T\Delta S$$

A reaction is spontaneous when:

$$\Delta G < 0$$

Now check each combination.

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1. Case (A): $\Delta H > 0,\; \Delta S < 0$

Then

$$\Delta G = (+) - T(-) = (+) + (+)$$

So $\Delta G$ is always positive for any $T$. Hence the reaction is non-spontaneous at any temperature.

So (A) is correct.

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2. Case (B): $\Delta H > 0,\; \Delta S > 0$

Then

$$\Delta G = (+) - T(+)$$

At low $T$, the $\Delta H$ term dominates, so $\Delta G > 0$. At high $T$, the $T\Delta S$ term can dominate, so $\Delta G < 0$.

Therefore this reaction is spontaneous at high temperature, not low temperature.

So (B) is incorrect.

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3. Case (C): $\Delta H < 0,\; \Delta S < 0$

Then

$$\Delta G = (-) - T(-) = (-) + (+)$$

At low $T$, the negative $\Delta H$ term dominates, so $\Delta G < 0$. At high $T$, the positive $T|\Delta S|$ term can dominate, so $\Delta G > 0$.

Therefore the reaction is spontaneous at low temperature. The table says low $T$, non-spontaneous, which is wrong.

So (C) is incorrect.

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4. Case (D): $\Delta H < 0,\; \Delta S > 0$

Then

$$\Delta G = (-) - T(+)$$

Both terms make $\Delta G$ negative for all temperatures. So the reaction is spontaneous at any temperature.

So (D) is correct.

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5. Incorrect combinations

The incorrect rows are:

• (B)
• (C)

Hence the correct option is:

$$\boxed{\text{D: (B) and (C) only}}$$