Alright — let's do this. Step 1: Find the coefficient of $x^3$ in the expansion of $(1+x)^5$. The general term in the binomial expansion of $(a+b)^n$ is given by $T_{k+1} = \binom{n}{k} a^{n-k} b^k$. For $(1+x)^5$, we have $a=1$, $b=x$, and $n=5$. We want the term with $x^3$, so we set $k=3$. $$T_{3+1} = \binom{5}{3} (1)^{5-3} (x)^3$$ $$T_4 = \binom{5}{3} (1)^2 (x)^3$$ Calculate $\binom{5}{3}$: $$\binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5!}{3!2!} = \frac{5 \times 4}{2 \times 1} = 10$$ So the term is $10 \cdot 1 \cdot x^3 = 10x^3$. The coefficient of $x^3$ is $10$. The coefficient of $x^3$ is $\boxed{\text{10}}$. Step 2: Find the coefficient of $x^2$ in the expansion of $(2x-3)^4$. The general term in the binomial expansion of $(a+b)^n$ is given by $T_{k+1} = \binom{n}{k} a^{n-k} b^k$. For $(2x-3)^4$, we have $a=2x$, $b=-3$, and $n=4$. $$T_{k+1} = \binom{4}{k} (2x)^{4-k} (-3)^k$$ We want the term with $x^2$, so the power of $x$ must be $2$. From $(2x)^{4-k}$, the power of $x$ is $4-k$. Set $4-k=2$: $$4-k=2 \implies k=2$$ Now substitute $k=2$ into the general term: $$T_{2+1} = \binom{4}{2} (2x)^{4-2} (-3)^2$$ $$T_3 = \binom{4}{2} (2x)^2 (-3)^2$$ Calculate $\binom{4}{2}$: $$\binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} = \frac{4 \times 3}{2 \times 1} = 6$$ Substitute the values: $$T_3 = 6 \cdot (4x^2) \cdot (9)$$ $$T_3 = 6 \cdot 4 \cdot 9 \cdot x^2$$ $$T_3 = 216x^2$$ The coefficient of $x^2$ is $216$. The coefficient of $x^2$ is $\boxed{\text{216}}$. Step 3: Find the term independent of $x$ in the expansion of $(1-2x)$. The expression given is $(1-2x)$. The term independent of $x$ is the constant term in this expression. The constant term is $1$. The term independent of $x$ is $\boxed{\text{1}}$. Step 4: Find the sum of coefficients in the expansion of $(a-b)^n$. To find the sum of coefficients in any polynomial expansion, substitute $1$ for each variable. In the expansion of $(a-b)^n$, we substitute $a=1$ and $b=1$. Sum of coefficients $= (1-1)^n = 0^n$. If $n > 0$, then $0^n = 0$. If $n=0$, then $(a-b)^0 = 1$, and the sum of coefficients is $1$. Assuming $n$ is a positive integer, which is typical for binomial expansion questions of this type. The sum of coefficients is $\boxed{\text{0}}$. Step 5: Find the middle term in the expansion of $(1-2x)^{-4}$. The expression $(1-2x)^{-4}$ represents an infinite series expansion, as the exponent is a negative integer. For example, using the generalized binomial theorem $(1+y)^n = 1 + ny + \frac{n(n-1)}{2!}y^2 + \dots$ Here, $y = -2x$ and $n = -4$. $$(1-2x)^{-4} = 1 + (-4)(-2x) + \frac{(-4)(-4-1)}{2!}(-2x)^2 + \dots$$ $$= 1 + 8x + \frac{(-4)(-5)}{2}(4x^2) + \dots$$ $$= 1 + 8x + 10(4x^2) + \dots$$ $$= 1 + 8x + 40x^2 + \dots$$ An infinite series does not have a "middle term" because there is no finite number of terms to determine a central position. The expansion of $(1-2x)^{-4}$ is an infinite series and therefore has $\boxed{\text{no middle term}}$.