What you basically have to determine is which modes are controllable and observable. Checking the rank of the controllability and observability matrices won't tell you which modes are controllable and observable, only whether the entire system is minimal (and to some extend how many modes are not present). A better method in this case would be the Hautus lemma. However a direct application of this would this require you to check the rank of eight matrices (all four eigenvalues with $B$ for controllability and $C$ for observability). This can be reduced significantly by using the similarity transformation $\hat{x} = V^{-1}\,x$, which gives
\begin{align}
\hat{A} &= V^{-1}\,A\,V = \Lambda, \\
\hat{B} &= V^{-1}\,B, \\
\hat{C} &= C\,V.
\end{align}
Here $\hat{B}$ is twice the first column plus the last column of $V^{-1}$ and $\hat{C}$ is the first minus the second row of $V$, which gives
$$
\hat{B} = \begin{bmatrix}
1 & 0 & 1 & 1
\end{bmatrix}^\top, \quad
\hat{C} = \begin{bmatrix}
0 & 0 & 1 & 1
\end{bmatrix}.
$$
From this result it is relatively easy to see, without actually applying the Hautus test, which modes are controllable and observable since $\hat{A}$ is diagonal, so the system can be seen as four decoupled first order systems. The first component of the state $\hat{x}$ has the associated eigenvalue $-4$ and its dynamics does get an input passed through by $\hat{B}$ (its first component is nonzero), but does not appear in the output (the first component of $\hat{C}$ is zero), so this state component is controllable but not observable. Similarly it can be shown that the state with eigenvalue $-2$ not controllable and not observable, and the states with eigenvalues $0$ and $1$ are both controllable and observable.
So a minimal representation of the system would only contain the states which are both controllable and observable. This would be the states with eigenvalues $0$ and $1$, therefore answer B) would be the only option. However B) is not minimal, since by factoring out the 2 it is easy to see that it has also a zero at $1$, which would mean you could do pole zero cancelation. This would imply the system was not minimal, but earlier we showed that it is minimal. So all that remains is E) none of the above.
The minimal transfer function can be calculated relatively easily by considering only the controllable and observable parts. A minimal state space model would then become
$$
\left[\begin{array}{c|c}
A_m & B_m \\ \hline C_m
\end{array}\right] =
\left[\begin{array}{cc|c}
0 & 0 & 1 \\ 0 & 1 & 1 \\ \hline 1 & 1
\end{array}\right]
$$
Its corresponding transfer function can be calculated with $C_m\,(s\,I-A_m)^{-1}B_m$, where $s\,I-A_m$ is diagonal so its inverse is just taking the inverse of each diagonal element. Using this inverse gives the following expression for the transfer function
$$
H(s) = \frac{1}{s} + \frac{1}{s-1} = \frac{2\,s-1}{s(s-1)}.
$$
Best Answer
Transfer functions do not describe the ratio of the output and the state, but the ratio of the output and the input of a given linear time invariant (LTI) system.
So you have $$ G(s)=\frac{Y(s)}{U(s)}=\frac{100(s-2)}{(s-3)(s+4)} $$ and so $$ \begin{align} Y(s)(s-3)(s+4)&=U(s)100(s-2)\\ Y(s)s^2+Y(s)s-12 Y(s)&=100 U(s)s-200 U(s) \end{align} $$ and so $$ \ddot{y}+\dot{y}-12y=100\dot{u}-200u $$ Now there are two issues: You have a second derivative and an input derivative which are both not present in the state space form $$ \begin{align} \dot{x}=Ax+Bu\\ y=Cx+Du \end{align} $$ First we solve the problem of the second derivative by introducing state variables: $$ \begin{align} z_1&=y\\ z_2&=\dot{y} \end{align} $$ So you get $$ \dot{z}_1=z_2,\dot{z}_2+z_2-12z_1=100\dot{u}-200u $$ Solving for the state derivatives: $$ \begin{align} \dot{z}_1&=z_2\\ \dot{z}_2&=12z_1-z_2+100\dot{u}-200u\\ y&=z_1 \end{align} $$ Now you only need to get rid of the input derivative. This is a question that has been answered several times on this site, for example here or here. You can do this by defining $$ \begin{align} x_1&=z_1+a_1u\\ x_2&=z_2+a_2u \end{align} $$ Now differentiate ... $$ \begin{align} \dot{x}_1&=\dot{z}_1+a_1\dot{u}\\ \dot{x}_2&=\dot{z}_2+a_2\dot{u} \end{align} $$ ... insert $\dot{z}_1,\dot{z}_2$ ... $$ \begin{align} \dot{x}_1&=z_2+a_1\dot{u}\\ \dot{x}_2&=12z_1-z_2+100\dot{u}-200u+a_2\dot{u}\\ y&=z_1 \end{align} $$ ... and replace $z_1$ with $x_1-a_1u$ and $z_2$ with $x_2-a_2u$ to get $$ \begin{align} \dot{x}_1&=x_2-a_2u+a_1\dot{u}\\ \dot{x}_2&=12(x_1-a_1u)-x_2+a_2u+100\dot{u}-200u+a_2\dot{u}\\ y&=x_1-a_1u \end{align} $$ We can see that $a_1$ is not helpful so set it to zero. However $a_2$ can be used to eliminate the $\dot{u}$ term by setting it to $-100$. So we have $a_1=0,a_2=-100$ and so $$ \begin{align} \dot{x}_1&=x_2+100u\\ \dot{x}_2&=12x_1-x_2-300u\\ y&=x_1 \end{align} $$ Or, using the matrix form $\dot{x}=Ax+Bu, y=Cx+Du$ with $$ A=\begin{pmatrix} 0&1\\12&-1 \end{pmatrix},B=\begin{pmatrix}100\\-300\end{pmatrix},C=\begin{pmatrix}1&0\end{pmatrix},D=0 $$ This is different to the stated solution only up to a similarity transformation $q=Tx$ with $$ \begin{pmatrix} q_1\\q_2 \end{pmatrix}=\underbrace{\begin{pmatrix} \frac{1}{50}&\frac{1}{300}\\ \frac{1}{200}&\frac{1}{600} \end{pmatrix}}_{T}\begin{pmatrix} x_1\\x_2 \end{pmatrix} $$ because the transformation used earlier is not unique. Apply this you get $\dot{q}=\hat{A}q+\hat{B}u,y=\hat{C}q+\hat{D}u$ with $$ \hat{A}=TAT^{-1}=\begin{pmatrix} -1&12\\1&0 \end{pmatrix},\hat{B}=TB=\begin{pmatrix}1\\0\end{pmatrix},\hat{C}=CT^{-1}\begin{pmatrix}100&-200\end{pmatrix},\hat{D}=D=0, $$ i.e. the solution stated in your question.