$\underline{\textbf{Given}}$
$T_0 = 1; \ T_1 = 2; \ T_n = 5T_{n-1} - 6T_{n-2} \text{ for } (n \ge 2)$
$\underline{\textbf{Observations}}$
$$\begin{align}
\begin{pmatrix}T_n \\ T_{n-1}\end{pmatrix} &= \begin{pmatrix}5 & -6 \\ 1 & 0\end{pmatrix}\begin{pmatrix} T_{n-1}\\ T_{n-2}\end{pmatrix} \\
&= \begin{pmatrix}5 & -6 \\ 1 & 0\end{pmatrix}\begin{pmatrix}5 & -6 \\ 1 & 0\end{pmatrix}\begin{pmatrix} T_{n-2}\\ T_{n-3}\end{pmatrix} \\
&= \begin{pmatrix}5 & -6 \\ 1 & 0\end{pmatrix}^{n-1}\begin{pmatrix} T_{n-(n-1)}\\ T_{n-1-(n-1)}\end{pmatrix} = \underbrace{\begin{pmatrix}5 & -6 \\ 1 & 0\end{pmatrix}^{n-1}}_{A^{n-1}}\underbrace{\begin{pmatrix} 2 \\ 1\end{pmatrix}}_{F} \\
\end{align}$$
$A$ is invertible $2\times 2$ matrix. Hence, we can use the power of diagonalization to solve for $A^{n-1}$ neatly.
$\underline{\textbf{Diagonalization}}$
$\Lambda = P^{-1}AP$ or $A = P\Lambda P^{-1}$, where $\Lambda$ is $2\times 2$ diagonal matrix and $\Lambda_{i,i} = \lambda_i$, here $\lambda_i$ is eigen value of $A$. $P$ is $2\times 2$ invertible matrix such that $P = \begin{pmatrix}e_1 & e_2 \end{pmatrix}$, where $e_i$ is eigenvector corresponding to $\lambda_i$. Hence, using diagonalization we can represent any matrix in terms of its eigenvalues and eigenvectors.
$A^m = \underbrace{(P\Lambda P^{-1})(P\Lambda P^{-1})\dots(P\Lambda P^{-1})}_{m \text{ times}} = P\Lambda^m P^{-1} \implies \boxed{A^{n-1}F = P \Lambda^{n-1}P^{-1}F}$
Eigen values of $A$ are $\lambda_1 = 3$, $\lambda_2 = 2$, so $\Lambda = \begin{pmatrix}3 & 0 \\ 0 & 2 \end{pmatrix}$. $e_1 = \begin{pmatrix}3 \\ 1 \end{pmatrix}$ and $e_2 = \begin{pmatrix}2 \\ 1 \end{pmatrix}$, so $P = \begin{pmatrix}3 & 2 \\ 1 & 1 \end{pmatrix}$ and $P^{-1} = \begin{pmatrix}1 & -2 \\ -1 & 3 \end{pmatrix}$
$$\begin{align}
A^{n-1}F &= \begin{pmatrix}3 & 2 \\ 1 & 1 \end{pmatrix} \begin{pmatrix}3 & 0 \\ 0 & 2 \end{pmatrix} ^{n-1} \begin{pmatrix}1 & -2 \\ -1 & 3 \end{pmatrix} \begin{pmatrix} 2 \\ 1\end{pmatrix} \\
&= \begin{pmatrix}3 & 2 \\ 1 & 1 \end{pmatrix} \begin{pmatrix}3^{n-1} & 0 \\ 0 & 2^{n-1} \end{pmatrix} \begin{pmatrix}1 & -2 \\ -1 & 3 \end{pmatrix} \begin{pmatrix} 2 \\ 1\end{pmatrix} \\
&= \begin{pmatrix}3^{n} & 2^{n} \\ 3^{n-1} & 2^{n-1} \end{pmatrix} \begin{pmatrix} 0 \\ 1\end{pmatrix} \\
A^{n-1}F &= \begin{pmatrix} 2^n \\ 2^{n-1} \end{pmatrix} \\
\end{align}$$
$\underline{\textbf{Final Solution}}$
$\begin{pmatrix}T_n \\ T_{n-1}\end{pmatrix} = A^{n-1}F = \begin{pmatrix} 2^n \\ 2^{n-1} \end{pmatrix} \implies T_n = 2^n$. Hence $\boxed{T_n = \Theta(2^n)}$.
$\textbf{Option (A) is correct}$