Choose the correct statement(s) among the following.

The problem asks which of the given statements are correct, so we must check each one separately.

Statement A: $$SnCl_2\cdot 2H_2O$$ is a reducing agent.
Tin is in the +2 oxidation state in $$Sn^{2+}$$. $$Sn^{2+}$$ can easily lose two more electrons to reach the stable +4 state, therefore it behaves as a reducing agent (for example, it reduces $$Fe^{3+} \rightarrow Fe^{2+}$$ in qualitative analysis).
Hence Statement A is correct.

Statement B: $$SnO_2$$ reacts with $$KOH$$ to form $$K_2[Sn(OH)_6]$$.
$$SnO_2$$ is amphoteric. With a strong base it gives the stannate ion $$[Sn(OH)_6]^{2-}$$:
$$SnO_2 + 2\,KOH + 2\,H_2O \;\rightarrow\; K_2[Sn(OH)_6]$$
Therefore Statement B is correct.

Statement C: A solution of $$PbCl_2$$ in $$HCl$$ contains $$Pb^{2+}$$ and $$Cl^{-}$$ ions.
In excess chloride ion, sparingly soluble $$PbCl_2$$ forms the soluble complex $$[PbCl_4]^{2-}$$:
$$PbCl_2(s) + 2\,Cl^- \;\rightleftharpoons\; [PbCl_4]^{2-}$$
Hence free $$Pb^{2+}$$ is essentially absent; Statement C is wrong.

Statement D: The reaction of $$Pb_3O_4$$ with hot dilute $$HNO_3$$ to give $$PbO_2$$ is a redox reaction.
$$Pb_3O_4$$ consists of two $$Pb^{2+}$$ ions and one $$Pb^{4+}$$ ion (it may be written $$2\,PbO \cdot PbO_2$$). When it reacts with dilute $$HNO_3$$, the typical equation is
$$Pb_3O_4 + 4\,HNO_3 \;\rightarrow\; PbO_2 + 2\,Pb(NO_3)_2 + 2\,H_2O$$
The oxidation states remain: two Pb stay at +2 (in $$Pb(NO_3)_2$$) and one Pb stays at +4 (in $$PbO_2$$). No element experiences an increase and another a decrease in oxidation state simultaneously, so no redox change takes place. Statement D is wrong.

Thus only Statements A and B are correct.

Option A (SnCl2·2H2O is a reducing agent), Option B (SnO2 gives K2[Sn(OH)6] with KOH)