The best way to approach such a question is by checking the solubility rules for each of the ionic compounds (more here: http://www.chem.sc.edu/faculty/morgan/resources/solubility/).
Let's start with #Co(NO_3)_2#. According to the aforementioned solubility rules, all nitrate (#NO_3^-#) salts are soluble, so we would get
#Co(NO_3)_(2(aq)) -> Co^(2+)(aq) + 2NO_3^(-)(aq)#
This means that the concentrations of #Co^(2+)# and #NO_3^-# will be
#(0.50 mol es Co(NO_3)_2)/(1L solution) * (1 mol e Co^(2+))/(1 mol e Co(NO_3)_2) = 0.50 M# #Co^(2+)#
#(0.50 mol es Co(NO_3)_2)/(1L solution) * (2 mol es NO_3^(-))/(1 mol e Co(NO_3)_2) = 1.0 M# #NO_3^-#
Notice that the #NO_3^-#'s concentration will be twice as big as the concentration of #Co(NO_3)_2#, since 2 moles of the former are produced for every 1 mole of the latter.
Moving on to #Fe(ClO_4)_3#. The solubility rules tell us that all perchlorate (#ClO_4^-#) salts are soluble, so we will get
#Fe(ClO_4)_(3(aq)) -> Fe^(3+)(aq) + 3ClO_4^(-)(aq)#
Therefore, the concentrations of the two ions will be
#(1 mol e Fe(ClO_4)_3)/(1Lsolution) * (1 mol e Fe^(3+))/(1 mol e Fe(ClO_4)_3) = 1M# #Fe^(3+)#
#(1 mol e Fe(ClO_4)_3)/(1Lsolution) * (3 mol es ClO_4^-)/(1 mol e Fe(ClO_4)_3) = 3M# #ClO_4^-#
