Reduction of molality of methanol
If the molality is $1.44 \ce{m}$, it means that, there is 1.44 moles of methanol in $1 \ce{kg}$ of water.

To prepare this solution, we add the mass of  $1.44\times 32= 46.08\ce{g}$ of methanol to $1000\ce{g}$ of water.

The mass of the final solution is $1000+46.08= 1046.08\ce{g} $.

Here, you have 1 kg of this solution (water + methanol). So, it contains $\frac {46.08 \times 1000}{1046.08}=44.05\ce{g}$ of methanol. 

The mass of water in this one kg-solution is $1000-44.05=955.95\ce{g}$

The objective of the problem is to prepare a solution of molality $1 \ce{m}$,

i.e, it is prepared by adding $32\ce{g}$ of methaonl to $1000\ce{g}$ water.

It can also be prepared by adding $44.05\ce{g}$ of methanol to $\frac {44.05 \times 1000}{32}=1376.56
\ce{g}$ of water.

Our solution already contains $955.95\ce{g}$ of water.

So, we need to add only $1376.56-955.95=420.61\ce{g}$ of water.