Yes, the lower triangular matrices $L$ with 1’s on the diagonal form a group. Clearly, the product of two is a third. Further, the Gauss-Jordan method shows that the inverse of one is another.

No, the symmetric matrices do not form a group. For example, here are two symmetric matrices $A$ and $B$ whose product $A B$ is not symmetric.
$$
A=\left[\begin{array}{lll}
0 & 1 & 0 \
1 & 0 & 0 \
0 & 0 & 1
\end{array}\right], \quad B=\left[\begin{array}{lll}
1 & 2 & 3 \
2 & 4 & 5 \
3 & 5 & 6
\end{array}\right], \quad A B=\left[\begin{array}{lll}
2 & 4 & 5 \
1 & 2 & 3 \
3 & 5 & 6
\end{array}\right]
$$
No, the positive matrices do not form a group. For example, $\left(\begin{array}{ll}1 & 1 \ 0 & 1\end{array}\right)$ is positive, but its inverse $\left(\begin{array}{rr}1 & -1 \ 0 & 1\end{array}\right)$ is not.
Yes, clearly, the diagonal invertible matrices form a group.
Yes, clearly, the permutation matrices matrices form a group.
Yes, the matrices with $Q^{\mathrm{T}}=Q^{-1}$ form a group. Indeed, if $A$ and $B$ are two such matrices, then so are $A B$ and $A^{-1}$, as
$$
(A B)^{\mathrm{T}}=B^{\mathrm{T}} A^{\mathrm{T}}=B^{-1} A^{-1}=(A B)^{-1} \quad \text { and }\left(A^{-1}\right)^{\mathrm{T}}=\left(A^{\mathrm{T}}\right)^{-1}=A^{-1} .
$$
There are many more matrix groups. For example, given two, the block matrices $\left(\begin{array}{cc}A & 0 \ 0 & B\end{array}\right)$ form a third as $A$ ranges over the first group and $B$ ranges over the second. Another example is the set of all products $c P$ where $c$ is a nonzero scalar and $P$ is a permutation matrix of given size.

(a) Let $\mathbf{s}, \mathbf{s}^{\prime}$ be vectors in $\mathbf{S}$, let $\mathbf{t}, \mathbf{t}^{\prime}$ be vectors in $\mathbf{T}$, and let $c$ be a scalar. Then
$$
(\mathbf{s}+\mathbf{t})+\left(\mathbf{s}^{\prime}+\mathbf{t}^{\prime}\right)=\left(\mathbf{s}+\mathbf{s}^{\prime}\right)+\left(\mathbf{t}+\mathbf{t}^{\prime}\right) \quad \text { and } \quad c(\mathbf{s}+\mathbf{t})=c \mathbf{s}+c \mathbf{t}
$$
Thus $\mathbf{S}+\mathbf{T}$ is closed under addition and scalar multiplication; in other words, it satisfies the two requirements for a vector space.
(b) If $\mathbf{S}$ and $\mathbf{T}$ are distinct lines, then $\mathbf{S}+\mathbf{T}$ is a plane, whereas $\mathbf{S} \cup \mathbf{T}$ is not even closed under addition. The span of $\mathbf{S} \cup \mathbf{T}$ is the set of all combinations of vectors in this union. In particular, it contains all sums $\mathbf{s}+\mathbf{t}$ of a vector $\mathbf{s}$ in $\mathbf{S}$ and a vector $\mathbf{t}$ in $\mathbf{T}$, and these sums form $\mathbf{S}+\mathbf{T}$. On the other hand, $\mathbf{S}+\mathbf{T}$ contains both $\mathbf{S}$ and $\mathbf{T}$; so it contains $\mathbf{S} \cup \mathbf{T}$. Further, $\mathbf{S}+\mathbf{T}$ is a vector space. So it contains all combinations of vectors in itself; in particular, it contains the span of $\mathbf{S} \cup \mathbf{T}$. Thus the span of $\mathbf{S} \cup \mathbf{T}$ is $\mathbf{S}+\mathbf{T}$.

Each column of $A B$ is a combination of the columns of $A$ (the combining coefficients are the entries in the corresponding column of $B$ ). So any combination of the columns of $[A A B]$ is a combination of the columns of $A$ alone. Thus $A$ and $[A A B]$ have the same column space.
Let $A=\left(\begin{array}{ll}0 & 1 \ 0 & 0\end{array}\right)$. Then $A^2=0$, so $\mathbf{C}\left(A^2\right)=\mathbf{Z}$. But $\mathbf{C}(A)$ is the line through $\left(\begin{array}{l}1 \ 0\end{array}\right)$.
An $n$ by $n$ matrix has $\mathbf{C}(A)=\mathbf{R}^n$ exactly when $A$ is an invertible matrix, because $A x=b$ is solvable for any given $b$ exactly when $A$ is invertible.