To neutralize 1.65g lioh, how much .150 m hcl would be needed?

The reaction between LiOH and HCl is;   
LiOH + HCl → LiCl + H₂O

The stoichiometric ratio between LiOH and HCl is 1 : 1
moles of LiOH added = moles of HCl needed to neutralize.

Molar mass of LiOH = 24 g/mol

moles = mass / molar mass
LIOH moles  = 1.65 / 24 = 0.06875 mol
Hence needed HCl moles = 0.06875 mol

Molarity = moles (mol) / Volume (L)
Hence needed HCl volume = moles / molarity                                           
                                            = 0.06875 mol / 0.150 mol/L                                           
                                            = 0.458 L = 458 mL

You know how the first one if filled up like the objects do the same but with the other categories