Kohlrausch's Law

By Kohlrausch's law: $\Lambda^\circ_{m(CH_3COOH)} = \Lambda^\circ_{m(CH_3COONa)} + \Lambda^\circ_{m(HCl)} - \Lambda^\circ_{m(NaCl)}$ $= 91.0 + 426.16 - 126.45 = 390.71\ S\ cm^2\ mol^{-1}$

By Kohlrausch's law, $\lambda^0_{CH_3COOH}=\lambda^0_{CH_3COO^-}+\lambda^0_{H^+}=\lambda^0_{CH_3COOK}+\frac{1}{2}\lambda^0_{H_2SO_4}-\frac{1}{2}\lambda^0_{K_2SO_4}=z+\frac{x}{2}-\frac{y}{2}=\frac{(x-y)}{2}+z$.

Kohlrausch's law: at infinite dilution each ion makes its own contribution to the conductance, independent of the other ion.
In terms of equivalent conductances, one equivalent of any electrolyte contains one equivalent of the cation and one equivalent of the anion.
So $\Lambda^\circ_{eq}(\text{electrolyte}) = \Lambda^\circ_{eq}(\text{cation}) + \Lambda^\circ_{eq}(\text{anion})$, with no stoichiometric multipliers.
The ionic values given are already equivalent conductances, so: $\Lambda^\circ_{eq}[Al_2(SO_4)_3] = \Lambda^\circ_{Al^{3+}} + \Lambda^\circ_{SO_4^{2-}}$
(The form $2\Lambda^\circ + 3\Lambda^\circ$ would apply only if molar ionic conductances were given and the molar conductance were asked.)

By Kohlrausch's law, the limiting equivalent conductance of an electrolyte is the sum of the limiting equivalent conductances of its ions.
The value 127 for barium corresponds to the whole $Ba^{2+}$ ion; per equivalent it is $\dfrac{1}{2} \times 127 = 63.5$.
$\Lambda^\circ_{eq}(BaCl_2) = \lambda^\circ\left(\dfrac{1}{2}Ba^{2+}\right) + \lambda^\circ(Cl^-) = 63.5 + 76$
$= 139.5$ ohm$^{-1}$ cm$^2$ eq$^{-1}$