I am having problem with this question including topics of redox reactions and charges
As the formula of the substance is known, you should first state the formula of the ions produced when the substance is dissolved into water. Here, $\ce{K_4Fe(CN)_6}$ gets dissolved in water and produced $4$ ions $\ce{K^+}$ so that the $4$ corresponding negative charges  must be fixed on the remaining anion, which has the formula $\ce{[Fe(CN)_6]^{4-}}$ with $4$ negative charges. Now you have to think how to synthesize this complex anion from $\ce{1 Fe^x+}$ ion and $\ce{6 CN^-}$ ions. The equation is $$\ce{Fe^{x+} + 6 CN^-  -&gt; [Fe(CN)_6]^{4-}}$$ Counting the charges gives x positive charges from $\ce{Fe^{x+}}$, 6 negative charges form the six cyanide ions, and this makes a total of $\ce{4-}$. So the initial $\ce{Fe}$ ion must have been charged $+2$
The nature of the ion $\ce{CN^-}$ remains to be explained : Well ! The carbon atom has 4 independent valence electrons. Nitrogen has 5 valence electrons, but 2 of them are taken in a doublet. So there is only 3 available electrons in N atoms for bonding. When 1 C is attached to 1 N, this makes 3 covalent bonds between C and N. As a consequence one electron is still available on the carbon atom. This is not a stable compound. The carbon atom needs one more electron to get 8 electrons in its outer layer. This electron can be offered by one H atom, which would make a covalent molecule HCN, with one covalence between H and C, and three covalences between C and N. But this last electron can also be given by a sodium Na or a potassium atom K. This would produce the compound NaCN or KCN, where  Na (or K) and the carbon atom are bound in a ionic bond between $\ce{Na^+}$ (or $\ce{K^+}$) and the ion $\ce{CN^-}$. This is the origin of the ion cyanide $\ce{CN^-}$.