1. We need to distinguish nitrate salts of $Mn^{2+}$ and $Cu^{2+}$ using the given statements.
   So we test each option independently.

2. Option A: $Mn^{2+}$ shows the characteristic green color in the flame test

• Flame test colors are due to excitation of electrons.
• $Cu^{2+}$ salts characteristically give a blue-green flame.
• $Mn^{2+}$ does not give a characteristic green flame in ordinary flame test.

Therefore, Option A is false.

3. Option B: Only $Cu^{2+}$ shows the formation of precipitate by passing $H_2S$ in acidic medium

• In qualitative analysis, $H_2S$ in acidic medium precipitates only those metal sulfides whose solubility product is very low.
• $Cu^{2+}$ forms black precipitate of copper sulfide: $Cu^{2+} + H_2S \rightarrow CuS\downarrow + 2H^+$
• $Mn^{2+}$ does not precipitate in acidic medium because $MnS$ is not sufficiently insoluble under these conditions.

Therefore, only $Cu^{2+}$ gives precipitate in acidic medium.
So, Option B is true.

4. Option C: Only $Mn^{2+}$ shows the formation of precipitate by passing $H_2S$ in faintly basic medium

• In faintly basic medium, sulfide ion concentration increases.
• $Mn^{2+}$ forms flesh-colored precipitate of $MnS$: $Mn^{2+} + S^{2-} \rightarrow MnS\downarrow$
• But $Cu^{2+}$ also forms $CuS$ precipitate, and in fact it precipitates even in acidic medium itself.

Hence it is wrong to say only $Mn^{2+}$ shows precipitate in faintly basic medium.

Therefore, Option C is false.

5. Option D: $Cu^{2+}/Cu$ has higher reduction potential than $Mn^{2+}/Mn$

Standard reduction potentials are: $E^\circ(Cu^{2+}/Cu)=+0.34\,V$ $E^\circ(Mn^{2+}/Mn)=-1.18\,V$

Clearly, $+0.34\,V > -1.18\,V$

Therefore, $Cu^{2+}/Cu$ has higher reduction potential.

So, Option D is true.

6. Final selection

• A: False
• B: True
• C: False
• D: True

Hence, the correct options are: $\boxed{B, D}$

7. Comparison with stored correct answer

Stored correct answer: B, D
Derived answer: B, D

They match exactly.