What are the concentrations of barium and sulfate ions in this solution?

You start with 75mL 0.030M BaCl2, which means you have 0.030 mol Ba/1000mL*75mL=0.00225 mol Ba.  You have 0.040mol SO4/1000mL*135mL=0.0054 mol SO4.  When those 2 get together, it forms insoluble BaSO4 (technically, a very tiny amt is soluble, but it is so low that it is beyond the range of your sig figs).  So the 0.00225 mol Ba combines with 0.00225 mol SO4 and precipitates leaving 0.0054-0.00225=0.00315 mol SO4 in solution.  That is in 210mLs of solution (75+135) so the molarity of SO4 is 0.00315*1000/210=0.015M.  There is "zero" Ba in solution (actually somewhere around 0.00001M)

0.03M x 0.075L = 0.00225moles BaCl2.....0.00225moles Ba 2+, 0.0045moles Cl-
0.04M x 0.135L = 0.0054moles K2SO4.....0.0108moles K+, 0.0054 moles SO4 2-

0.00225moles Ba 2+ + 0.0054moles SO4 2- 0.00225moles BaSO4 + 0.00315moles SO4 2-
initial [BaSO4] = 0.0107M
initial [SO4 2-] = 0.00315 / 0.21L = 0.015M

Ksp of BaSO4 = 1.1x10^-10 = [Ba 2+][SO4 2-] / [BaSO4]
1.1x10^-10 = (x)(0.015+x) / 0.0107 - x
-1.1x10^-10x + 1.18x10^-12 = 0.015x + x^2
x^2 + 0.015x - 1.18x10^-12 = 0
no additional dissociation of BaSO4

final [Ba 2+] = 0, final [SO4 2-] = 0.015M