\begin{align*} T_{r + 1} & = {12 \choose r} (2x)^{12 - r} \left(3 \over x\right)^r \\ & = {12 \choose r} (2)^{12 - r} (x)^{12 - r} (3)^r \left(1 \over x\right)^r \phantom{0000000} [ (a \times b)^m = a^m \times b^m] \\ & = {12 \choose r} (2^{12 - r}) (x^{12 - r}) (3^r) (x^{-1})^r \phantom{0000000} \left[ a^{-n} = {1 \over a^n} \right] \\ & = {12 \choose r} (2^{12 - r}) (x^{12 - r}) (3^r) (x^{-r}) \phantom{00000000} [ (a^m)^n = a^{mn}] \\ & = {12 \choose r} (2^{12 - r}) (3^r) (x^{12 - r + (-r)}) \phantom{000000000} [a^m \times a^n = a^{m + n}] \\ & = {12 \choose r} (2^{12 - r}) (3^r) (x^{12 - 2r}) \\ \\ \text{Power of } & x = 12 - 2r \end{align*}

\begin{align} {1 \over x^8} & = x^{-8} \\ \\ \text{Let } 12 - 2r & = -8 \\ -2r & = -20 \\ r & = {-20 \over -2} \\ r & = 10 \\ \\ T_{10 + 1} & = {12 \choose 10} (2^{12 - 10}) (3^{10}) (x^{12 - 2(10)}) \\ & = (66)(2^2)(3^{10})(x^{-8}) \\ & = 15 \phantom{.} 588 \phantom{.} 936 x^{-8} \\ & = 15 \phantom{.} 588 \phantom{.} 936 \left(1 \over x^8\right) \\ \\ \text{Coefficient} & = 15 \phantom{.} 588 \phantom{.} 936 \end{align}

\begin{align} \text{Let } 12 - 2r & = 0 \\ -2r & = -12 \\ r & = {-12 \over -2} \\ r & = 6 \\ \\ T_{6 + 1} & = {12 \choose 6} (2^{12 - 6}) (3^{6}) (x^{12 - 2(6)}) \\ & = (924)(2^6)(3^6)(x^0) \\ & = 43 \phantom{.} 110 \phantom{.} 144 \end{align}

\begin{align} T_{r + 1} & = {15 \choose r} (x)^{15 - r} \left(-{1 \over 2x^8}\right)^{r} \\ & = {15 \choose r} (x^{15 - r}) \left(-{1 \over 2}\right)^r \left(1 \over x^8\right)^r \\ & = {15 \choose r} (x^{15 - r}) \left(-{1 \over 2}\right)^r (x^{-8})^r \\ & = {15 \choose r} (x^{15 - r}) \left(-{1 \over 2}\right)^r (x^{-8r}) \\ & = {15 \choose r} \left(-{1 \over 2}\right)^r (x^{15 - r + (-8r)}) \\ & = {15 \choose r} \left(-{1 \over 2}\right)^r (x^{15- 9r}) \\ \\ \text{Let } 15- 9r & = 0 \\ -9r & = -15 \\ r & = {-15 \over -9} \\ r & = {5 \over 3} \\ \\ \text{Since } r \text{ is not} & \text{ an integer, there is no term that is independent of } x \end{align}

\begin{align} T_{r + 1} & = {10 \choose r} (2x)^{10 - r} \left(-{1 \over x}\right)^r \\ \\ T_{8 + 1} & = {10 \choose 8} (2x)^{10 - 8} \left(-{1 \over x}\right)^8 \\ T_{9} & = (45)(2x)^2 \left(1 \over x^8\right) \\ & = (45)(4x^2)\left(1 \over x^8\right) \\ & = {180x^2 \over x^8} \\ & = {180 \over x^{6}} \\ & = 180 x^{-6} \end{align}

\begin{align} T_{r + 1} & = {8 \choose r} (x)^{8 - r} \left(-{2 \over x^2}\right)^r \\ \\ T_{4 + 1} & = {8 \choose 4} (x)^{8 - 4} \left(-{2 \over x^2}\right)^4 \phantom{000000} [9 \text{ terms in total so } 5 \text{th term is in the middle}] \\ T_5 & = (70)(x^4)\left(16 \over x^8\right) \\ & = {1120x^4 \over x^8} \\ & = {1120 \over x^4} \\ & = 1120 x^{-4} \end{align}