\(a) \,2i(3 + i)(2 + 4i) =2i(6+14i+4i^2) \\= 2i(2 + 14i)=4i+28i^2 = -28 + 4i.\)
b) \( \dfrac{(1+i)^{2}(2i)^{3}}{-2+i}\) \( =\dfrac{2i(-8i)}{-2+i}=\dfrac{16(-2-i)}{5}=-\dfrac{32}{5}-\dfrac{16}{5}i.\)
\(c) \, 3 + 2i + (6 + i)(5 + i)=3+2i+30+11i+i^2 \\= 3 + 2i + 29 + 11i = 32 + 13i.\)
\(d) \, 4 - 3i + \dfrac{5+4i}{3+6i} = 4 - 3i + \dfrac{(5+4i)(3-6i)}{3^2+6^2} \\ = 4-3i + \dfrac{15-18i-24i^2}{45}= 4 - 3i + \dfrac{39}{45}-\dfrac{18}{45}i \\= (4 + \dfrac{39}{45}) - (3 + \dfrac{18}{45})i= \dfrac{73}{15}-\dfrac{17}{5}i.\)