Hello Shivam,
f(x)=x^{4} -8x^{3}+4x^{2}+4x+39
(I'm assuming this was supposed to say 4x rather than 4^x based on the context of the problem.)
1. f(3+2i)=(3+2i)^{4}-8(3+2i)^{3}+4(3+2i)^{2}+4(3+2i)+39
I'm going to do each term separately starting with the squared one to keep it simple.
4(3+2i)(3+2i)
=4(9+6i+6i+4i^{2})
Now remember that i^{2}=-1
=4(9+12i-4)
=4(5+12i)
=20+48i
8(3+2i)^{3}
=8(5+12i)(3+2i) Since we already determined that (3+2i)^{2}=(5+12i)
=8(15+10i+36i+24i^{2})
=8(15+46i-24)
=8(-9+46i)
=-72+368i
(3+2i)^{4}
=(5+12i)(5+12i)
=25+60i+60i+144i^{2}
=25+120i-144
=-119+120i
Back to the original:
f(3+2i)=-119+120i-72+368i+20+48i+12+8i+39
=-120+544i
Therefore a=-120 and b=544
2. f(x)=2x^{4}+5x^{3}+7x^{2}-x+41
f(-2-√3i)=2(-2-√3i)^{4}+5(-2-√3i)^{3}+7(-2-√3i)^{2}-(-2-√3i)+41
7(-2-√3i)(-2-√3i)
=7(4-2√3i-2√3i+3i^{2})
=7(4-4√3i-3)
=7(1-4√3i)
=7-28√3i
5(-2-√3i)^{3}
=5(1-4√3i)(-2-√3i)
=5(-2-√3i+8√3i+12i^{2})
=5(-2+7√3i-12)
=5(-14+7√3i)
=-70+35√3i
7(-2-√3i)^{4}
=7(1-4√3i)(1-4√3i)
=7(1-4√3i-4√3i-48i^{2})
=7(1-8√3i+48)
=7(49-8√3i)
=343-56√3i
f(-2-√3i)=343-56√3i-70+35√3i+7-28√3i+2+√3i+41
=323-48√3i
In a long operation like this, there are a lot of opportunities for human errors, so you would be wise to run through it twice.