How to write a balanced nuclear equation for the reactions?

Keep in mind that the atomic number ("the bottom #") always matches for the atom/element
1)
beta emission is typically understood to mean β- emission:
\({}^{210}_{82}\mathrm{Pb} -> {}^{210}_{83}\mathrm{Bi} + {}^0_{-1}\mathrm{{e^{-}}} + {^{-}}{{v}_{e}}\)

So the lead decays to Bismuth, an electron, and an antineutrino to conserve energy

2) an alpha particle is a Helium-4 atom:
\({}^{54}_{26}\mathrm{Fe} + {}^{4}_{2}\mathrm{He} -> 2 {}^1_{1}\mathrm{{p^{+}}} + {}^{56}_{26}\mathrm{Fe}\)

3)
\({}^{40}_{18}\mathrm{Fe} + {}^{A}_{Z}\mathrm{X} -> {}^{43}_{19}\mathrm{K} + {}^1_{1}\mathrm{{p^{+}}}\)

The right side sums to 44 and 20, so A must be 4 and Z must be 2:
\({}^{40}_{18}\mathrm{Ar} + {}^{4}_{2}\mathrm{He} -> {}^{43}_{19}\mathrm{K} + {}^1_{1}\mathrm{{p^{+}}}\)

The species, therefore, is Helium-4.

4)
\({}^{A}_{Z}\mathrm{X} + {}^{4}_{2}\mathrm{He} -> {}^{233}_{91}\mathrm{Pa}\)
233 - 4 = 229 and 91 - 2 = 89, therefore the species is Actinium-229: \({}^{229}_{89}\mathrm{Ac}\):
\({}^{229}_{89}\mathrm{Ac} + {}^{4}_{2}\mathrm{He} -> {}^{233}_{91}\mathrm{Pa}\)