A number #n# is divisible by #4# when the number formed by its two right digits is divisible by #4# and the same number #n# is divisible by #9# when the sum of its digits is divisible by #9#.
We have
#c_1->3 xx 10 + diamond = 4 xx k_1#
#c_2->1+2+square+3+diamond = 9 xx k_2#
here #{square,diamond} in {0,1,2,3,4,5,6,7,8,9}#
From #c_1-> diamond in {2,6}#
Choosing #diamond = 2# we have #c_2->6+2+square_1=9 xx k_3#
so #square_1=1#
Choosing #diamond = 6# we have #c_2->6+6+square_2=9 xx k_4#
so #square_2 = 6#
The solutions are #n_1 = 12132# and #n_2 = 12636#