Solution

Convert each number into decimal form.

$(142)_b=1\cdot b^2+4\cdot b+2$

$=b^2+4b+2$

Now,

$(112)_{b-2}=1\cdot (b-2)^2+1\cdot (b-2)+2$

$=(b-2)^2+(b-2)+2$

$=b^2-4b+4+b-2+2$

$=b^2-3b+4$

Also,

$(75)_8=7\cdot 8+5=56+5=61$

Now,

$(142)b+(112){b-2}=(75)_8$

$(b^2+4b+2)+(b^2-3b+4)=61$

$2b^2+b+6=61$

$2b^2+b-55=0$

$2b^2+11b-10b-55=0$

$b(2b+11)-5(2b+11)=0$

$(b-5)(2b+11)=0$

So,

$b=5$