To find the sum of coefficients, substitute 1 for each variable in the expression.

For $(a+b)^n$, substitute $a=1$ and $b=1$.
Sum of coefficients $= (1+1)^n = 2^n$.

The sum of the coefficients is $\boxed{2^n}$.
This matches option c) $2^n$.

$T_{k+1} = \binom{n}{k} a^{n-k} b^k$.
Here, $a=1$, $b=2x$, and $n=6$.
We want the coefficient of $x^2$, so we set $k=2$.

The coefficient of $x^2$ is 60.

The coefficient of $x^2$ is $\boxed{\text{60}}$.
This matches option b) 60.

$T_{k+1} = \binom{n}{k} a^{n-k} b^k$.
Here, $a=1$, $b=-x$, and $n=6$.
We want the coefficient of $x^4$, so we set $k=4$.

The coefficient of $x^4$ is 15.

The coefficient of $x^4$ is $\boxed{\text{15}}$.
This matches option c) 15.

$T_{k+1} = \binom{n}{k} a^{n-k} b^k$.
Here, $a=2$, $b=3x$, and $n=4$.
We want the coefficient of $x^2$, so we set $k=2$.

The coefficient of $x^2$ is 216.

The coefficient of $x^2$ is $\boxed{\text{216}}$.
Since 216 is not among options a) -72, b) -72, c) 36, d) -36, the correct option is e) None of the above.

Here, $y=2x$ and $n=-3$.
We need the term containing $x^2$, which corresponds to the term with $y^2$.

The term is $\frac{n(n-1)}{2!}y^2$.

The coefficient of $x^2$ is 24.

The coefficient of the term containing $x^2$ is $\boxed{\text{24}}$.
Since 24 is not among options a) -3, b) -6, c) 6, d) 3, the correct option is e) None of the above.

$T_{k+1} = \binom{n}{k} a^{n-k} b^k$.
Here, $a=1$, $b=-4x$, and $n=5$.
The term independent of $x$ is the term where the power of $x$ is 0. This occurs when $k=0$.

The term independent of $x$ is 1.

The term independent of $x$ is $\boxed{\text{1}}$.
Since 1 is not among options a) -40, b) 40, c) -10, d) 10, the correct option is e) None of the above.