Battery acid solution preparation from 98% ACS Sulfuric Acid
The answer begins with the general calculation of mixing solutions with various densities. If you are interested only in the particular numeric results, or if you are scared by algebra, , skip the theoretical part between the horizontal lines.

There is the mixing cross rule, deriving the ratio of mass of 2 mixed liquids from their mass percentage.
$$\frac { m_2 }{m_1} = \frac{|w_1 - w_0|}{|w_2 - w_0|}=k\tag{1}$$
respectively
$$m_2 = k\cdot m_1 \tag{2}$$
and
$$m_0 = m_1 + m_2 = m_1 + k \cdot m_1 =  m_1 \cdot \left(1 + k \right) \tag{3}$$
where
$m_1,m_2, m_0$ are respective masses of mixed solution 1 and 2 and  ofthe final solution 0
$w_1,w_2,w_0$ are respective mass percentages of mixed solutions 1 and 2 and the final solution 0
As we are interested rather in volumes, we involve solution densities:
$$V_1 = \frac{m_1 }{\rho_1}\tag{4}$$
$$V_2 = \frac{m_2 }{\rho_2} = \frac{m_1 }{\rho_2} \cdot k\tag{5}$$
$$V_0 = \frac{m_0 }{\rho_0} = \frac{m_1 \cdot \left(1 + k \right) }{\rho_0}\tag{6}$$
We have the given final volume $V_0$ of the final solution of the density $\rho_0$.
Then, for the solution 1:
$$m_1  = V_0 \cdot \rho_0 \cdot \frac{1}{\left(1 + k \right)} \tag{7}$$
respectively the volume:
$$V_1  = V_0 \cdot \frac{ \rho_0}{\rho_1 } \cdot \frac{1}{1 + k}\tag{8}$$
For the solution 2:
$$m_2  = m_1  \cdot k = V_0 \cdot \rho_0 \cdot \frac {k}{1+k} \tag{9} $$
respectively the volume:
$$V_2  = V_0 \cdot \frac{\rho_0}{\rho_2 } \cdot \frac {k}{1+k} \tag{10} $$

Applying the particular data:
Solution 0: 37% $\ce{H2SO4}$,The density  $\pu{1.267 g/mL}$,  target volume $V_0  = \pu{100 mL}$
Solution 1: 98% $\ce{H2SO4}$.The density  $\pu{1.8361 g/mL}$
Solution 2: water = 0% $\ce{H2SO4}$, the density $\pu{1.000 g/mL}$
From the equation (1):
$k = \frac{|98-37|}{|0-37|}=61/37 \simeq 1.649$
We plug in the data into equations (8) and (10)
For  98% $\ce{H2SO4}$:
$$V_1  = \pu{100 mL} \cdot \frac{ \pu{1.267 g/mL}}{\pu{1.8361 g/mL} } \cdot \frac{1}{1 + 1.649} \simeq \pu{26 mL}\tag{8a}$$
For water:
$$V_2  = \pu{100 mL} \cdot \frac{ \pu{1.267 g/mL}}{ \pu{1.000 g/mL}} \cdot \frac {1.649}{1+1.649} \simeq \pu{79 mL} \tag{10a} $$
Be aware volumes are not additive. After cooling down, the total initial volume shrinks from $\pu{105 mL}$ to $\pu{100 mL}$.
For subsequent dilutions, it is an advantage to keep some volume of the residual 37% solution and to do the dissolution in 2 steps.
First, dilute the calculated portion of 98% H2SO4 in this rest of 37% H2SO4. After cooling, pour this mixture to the calculated volume of water.
The amount of heat per dilution would be divided into these 2 steps.
DO NOT forget to pour acid into water, as not vice versa !! Pouring water into acid would very propably end
by a hot acid splash into your face, as mixing is very exothermic. Mix them carefully, slowly, continually mixing and cool it by a water bath.
Pour the acid along the glass rod and not directly. Wear a lab coat or at least an apron, gloves and shield if possible.
Be aware even tiny drops can cause after some days/weeks holes in your clothes.