= Solution
Let $K$ be the common normalization converting the quoted $N_i$ to photon numbers per parent decay. Apply <nuclear level population balance> to scheme A, starting with the top level:
$$
\begin{aligned}
\Gamma_3&=K(N_2+N_5+N_6),\\
\Gamma_2+KN_2&=K(N_3+N_4),\\
\Gamma_1+K(N_3+N_5)&=KN_1.
\end{aligned}
$$
For example, the $3\to2$ transition contributes $KN_2$ to the population of level 2; it must be subtracted from the total leaving level 2 to obtain its direct beta feeding. Similarly, a <photon> in the strong $1\to0$ line need not imply a direct beta transition to level 1.
The three equations give
$$
\Gamma_1=K(N_1-N_3-N_5),\quad
\Gamma_2=K(N_3+N_4-N_2),\quad
\Gamma_3=K(N_2+N_5+N_6).
$$
Adding them cancels the internal cascade counts:
$$
1-\Gamma_0=K(N_1+N_4+N_6),\qquad
K=\frac{1-\Gamma_0}{N_1+N_4+N_6}.
$$
This also has a simple physical meaning: each decay that initially feeds an excited daughter eventually contributes exactly one of the three transitions into the <ground state>. The number of all emitted <photons> is not the correct normalization, because a single cascade can emit several.
Consequently the required formulae are
$$
\boxed{\begin{aligned}
\Gamma_1&=(1-\Gamma_0)\frac{N_1-N_3-N_5}{N_1+N_4+N_6},\\
\Gamma_2&=(1-\Gamma_0)\frac{N_3+N_4-N_2}{N_1+N_4+N_6},\\
\Gamma_3&=(1-\Gamma_0)\frac{N_2+N_5+N_6}{N_1+N_4+N_6}.
\end{aligned}}
$$
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