1.
Let's look at the problem: the #"HCl"# in solution is #38%# by mass. If we assume a #100 "g"# sample, then then the mass of #"HCl"# in solution is #38"g"#. The number of moles of the solute is thus
#38cancel("g HCl")((1 "mol HCl")/(36.46cancel( "g HCl"))) = 1.04 "mol HCl"#
Now, let's use the density of the solution (#1.19 "g"/"mL"#), and the fact that we assumed a #100 "g"# sample to calculate the volume, in #"L"#:
#100 cancel("g")((1cancel( "mL"))/(1.19 cancel("g")))((1 "L")/(10^3 cancel("mL"))) = 0.0840 "L soln"#
The molarity of the solution is thus
#M = "mol solute"/"L soln" = (1.04 "mol HCl")/(0.0840 "L soln") = color(red)(12"mol"/"L"#
2.
In #1000"mL"# (or #1 "L"#) of a #1 M# solution, the number of moles of #"HCl"# is
#1 "mol"/cancel("L")(1 cancel("L")) = 1 "mol HCl"#
We'll use this number and the fact that #1.04 "mol HCl"# occupies a volume of #0.0840 "L"# to find the volume of solution needed:
#1 cancel("mol HCl")((0.0840 "L soln")/(1.04 cancel("mol HCl"))) = color(blue)(0.081 "L")#, or #color(blue)(81 "mL"#