Solution: We recall that a matrix $M$ of size $n \times n$ is diagonalizable if and only if $M$ has $n$ linearly independent eigenvectors. Let us find the eigenvalues of the matrix $M$. The characteristic polynomial of $M$ is given by
\begin{align*}
\det (M - \lambda I) & = \det \begin{bmatrix}
4 - \lambda & -3 \\
1 & -\lambda \\
\end{bmatrix} \\
& = \lambda ^2 - 4\lambda + 3.
\end{align*}
Eigenvalue of $M$ are the roots of the characteristic polynomial.
\[
\lambda ^2 - 4\lambda + 3 = (\lambda -3)(\lambda -1) = 0.
\]
Thus, the eigenvalues are $1$ and $3$. Since the eigenvalues are distinct, the corresponding eigenvectors will be linearly independent and hence the matrix $M$ is diagonalizable. Now we will prove the following lemma.
Let $M$ is a diagonalizable matrix and $p(x)$ be any polynomial. Then $f(M)$ is also diagonalizable.
Proof: Given that $M$ is diagonalizable matrix, so we can find nonsingular matrix $P$ and a diagonal matrix $D = \mathrm{diag}\left[ \lambda _1, \lambda _2, \dots, \lambda _n \right] $ such that
\[
M = P D P ^{-1}.
\]
Let
\[
f(x) = \sum_{i=0}^{k} a_i x^i, \quad a_k \neq 0,
\]
be any polynomial of degree $k$. Then
\begin{align*}
f(M) & = \sum_{i=0}^{k} a_i M^i \\
& = \sum_{i= 0} ^{k} a_i \left( P D P ^{-1} \right)^i \\
& = P \left[ \sum_{i=0}^{k} a_i D^i \right] P ^{-1} \\
& = P D' P ^{-1} ,
\end{align*}
where $D'$ is a diagonal matrix. Thus, $f(M)$ is diagonalizable.
Therefore, $M^8 + M^{12}$ and $M^7 + M^9$ are diagonalizable matrices. Thus, both P and Q are TRUE.