Evaluate option A: $SnCl_2\cdot 2H_2O$ is stannous chloride dihydrate, where tin is in the $+2$ oxidation state.

• $Sn^{2+}$ is easily oxidized to $Sn^{4+}$.
• Therefore, it acts as a reducing agent.

Hence, A is correct.

Evaluate option B: $SnO_2$ is amphoteric and reacts with strong bases like $KOH$.

The reaction is: $SnO_2 + 2KOH + 2H_2O \rightarrow K_2[Sn(OH)_6]$

So $SnO_2$ does react with $KOH$ to form hexahydroxostannate(IV), $K_2[Sn(OH)_6]$.

Hence, B is correct.

Evaluate option C: A solution of $PbCl_2$ in excess $HCl$ does not contain only $Pb^{2+}$ and $Cl^-$ ions.

• In chloride-rich medium, $Pb^{2+}$ forms complex chloro-species such as $[PbCl_3]^-$ and $[PbCl_4]^{2-}$.
• Therefore, the statement is incomplete/incorrect.

Hence, C is incorrect.

Evaluate option D: Consider the reaction of $Pb_3O_4$ with hot dilute $HNO_3$.

• $Pb_3O_4$ can be written as $2PbO\cdot PbO_2$, containing lead in both $+2$ and $+4$ states.
• On treatment with hot dilute nitric acid: $Pb_3O_4 + 4HNO_3 \rightarrow 2Pb(NO_3)_2 + PbO_2 + 2H_2O$
• Here, $Pb^{2+}$ remains $Pb^{2+}$ and $Pb^{4+}$ remains $Pb^{4+}$.
• No oxidation number changes occur.

Therefore, this is not a redox reaction.

Hence, D is incorrect.

Final answer: The correct statements are: $\boxed{A,\ B}$

Comparison with stored correct answer: Stored correct answer is A, B, which matches the derived answer.