$*)n_{OH^-}=n_{Na}+n_K\Rightarrow n_{OH^- min}=\frac{1,265}{39}<n_{OH^-}<\frac{1,265}{23}=n_{OH^- max} $$*)n_{Al(NO_3)_3}=0,015 mol$
$\Rightarrow \begin{cases}n_{OH^- min}<3n_{Al(NO_3)_3} \\ 3n_{Al(NO_3)_3}<n_{OH^- max}<4n_{Al(NO_3)_3} \end{cases}$
$\Rightarrow \begin{cases}m_{max}=0,015.78=1,17 gam \\ m_{min}=min[(\frac{1,265.78}{3.39});(0,015.4-\frac{1,265}{23}).78]=0,39 gam \end{cases}$
$\Rightarrow 0,39<m<1,17$