None of the conjectured inequalities hold.

This answer contains three counterexamples. The first one is to $(\star)$, while the second and third ones (below the line) refer to previous inequalities conjectured by the OP.

\begin{equation*}
A=\begin{bmatrix} 5 & 5\\\\ 5 & 5\end{bmatrix},\quad
B=\begin{bmatrix} 8 &4 \\\\ 4 & 2\end{bmatrix},\quad
C=\begin{bmatrix} 8 &6 \\\\ 6 & 5\end{bmatrix}.
\end{equation*}

Then $A^2+AB+AC$ is a rank-1 matrix, so its determinant is zero, while $A^2+AB+AC+BC=\begin{bmatrix}268 & 203\\\\ 224 & 169\end{bmatrix}$, so its determinant is $-180$.

The structure of this counterexample and of the other ones below is to setup matrices so that the left hand side becomes a rank-1 matrix, which will have determinant zero. Then, one can adjust the other terms to violate inequalities in any direction.

*EDIT* I'm adding explicit matrices where $A,B,C \succ 0$, but still we have a counterexample to quell the OP's insistence ;-)

\begin{equation*}
A=\begin{bmatrix}11&12&7\\\\ 12 & 14 & 8\\\\ 7 & 8 & 11\end{bmatrix},\
B=\begin{bmatrix}19 &14&7\\\\ 14& 14&6\\\\ 7&6&3\end{bmatrix},\
C=\begin{bmatrix}17&17&16\\\\ 17&19&17\\\\ 16&17&17\end{bmatrix}
\end{equation*}

For these matrices, $\det(A^2+AB+AC) \approx 2.35\times 10^5$, while $\det(A^2+AB+AC+BC) \approx 4.67 \times 10^4$.

A particularly cute counterexample for your last question (*edit*: where $\det(A+B) \ge \det(A)+\det(B)$ holds for non symmetric matrices with positive eigenvalues) is the following:

\begin{equation*}
A = \begin{bmatrix}
0.5 & 0 & 0& 0\\\\
1 & 0.5 & 0 & 0\\\\
1 & 1 & 0.5 & 0\\\\
1 & 1 & 1 & 0.5
\end{bmatrix},\quad B = A^T.
\end{equation*}

Then, $\det(A+B)=0$, but $\det(A)+\det(B) = 1/8$.

The updated question, whether $\det(XY+YZ) \ge \det(XY)+\det(YZ)$ holds is also false. Here is a nice counterexample.

\begin{eqnarray*}
X = \begin{bmatrix}
2 & 2\\\\
2 & 2
\end{bmatrix},\quad Y = \begin{bmatrix}
5 & 5\\\\
5 & 10
\end{bmatrix},\quad Z = \begin{bmatrix}
10 & 4\\\\
4 & 2
\end{bmatrix}.
\end{eqnarray*}
With this choice, $\det(XY+YZ) = -300$, while $\det(XY)+\det(YZ)=100$.

allthree matrices are positive definite, so that you feel more convinced ;-) – Suvrit May 27 '13 at 18:58