63 = (6!)/(3!(6-3)!) = (6!)/(3!3!) = (6 x 5 x 4)/(3 x 2 x 1) = 20

$T_{k+1} = \binom{n}{k} a^{n-k} b^k$.
Here, $a=1$, $b=x$, and $n=7$.
We want the coefficient of $x^4$, so we set $k=4$.

The coefficient of $x^4$ is 35.

The coefficient of $x^4$ is $\boxed{\text{35}}$.
This matches option b) 35.

$T_{k+1} = \binom{n}{k} a^{n-k} b^k$.
Here, $a=1$, $b=-5x$, and $n=4$.
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 $\boxed{\text{1}}$.
This is not among options a) -50, b) 50, c) -10, d) 10.
The correct option is e) None of the above.

For $(a+b)^n$, there are $n+1$ terms. Here $n=11$, so there are $11+1=12$ terms.

Since there are 12 terms (an even number), there are two middle terms. Their positions are $\frac{12}{2}=6$-th term and $\frac{12}{2}+1=7$-th term.

For $(1+x)^{11}$, $a=1$, $b=x$, $n=11$.
For the 6-th term, $k+1=6 \implies k=5$.

For the 7-th term, $k+1=7 \implies k=6$.

The middle terms are $\binom{11}{5} x^5$ and $\binom{11}{6} x^6$.
The options provided are powers of $x$. Option b) $x^5$ corresponds to the variable part of one of the middle terms.

The middle term (referring to the power of $x$) is $\boxed{x^5}$.
This matches option b) $x^5$.

For a matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the determinant is $ad-bc$.

Here, $a=3$, $b=4$, $c=2$, $d=5$.

The determinant is $\boxed{\text{7}}$.
This matches option b) 7.