With regards to block matrix multiplication, if you have a $m\times n$ matrix $A$ and a $n\times p$ matrix $B$ partitioned as $$A=\begin{pmatrix} \alpha_{a\times b} & \beta_{a\times (n-b)}\\ \gamma_{(m-a)\times b} & \delta_{(m-a)\times (n-b)} \end{pmatrix}, B=\begin{pmatrix} \eta_{b\times c} & \zeta_{b\times (p-c)} \\ \theta_{(n-b)\times c} & \mu_{(n-b)\times(p-c)} \end{pmatrix}$$ with $\alpha, \beta,\gamma,\delta,\eta,\zeta,\theta,\mu$ being submatrices of the above matrices with dimensions as shown above then the product matrix $AB$ is calculated exactly as if $\alpha,\beta,\gamma,\delta,\eta,\zeta,\theta,\mu$ were numbers (note that all the matrix multiplications below are compatible - i.e in a product, number of columns of the first matrix equals the number of rows in the second matrix) $$AB=\begin{pmatrix} \alpha\eta+\beta\theta &\alpha\zeta+\beta\mu \\ \gamma\eta+\delta\theta & \gamma\zeta+\delta\mu \end{pmatrix}$$
So we have $M_{3\times 3}=\begin{pmatrix} A_{2\times 2} & a_{2\times 1} \\ b_{1\times 2} & B\end{pmatrix}, M_{1}=\begin{pmatrix} A_{2\times 2} & 0_{2\times 1} \\ b_{1\times 2} & 1\end{pmatrix}$ where the respective dimensions of the matrices are written as subscript, with the $1\times 1$ matrix $B$ itself being a number only so $B_{1\times 1}=B$,
$a_{2\times 1}= (a_1 \quad a_2)^T, b_{1\times 2}=(b_1\quad b_2)$, and $0_{2\times 1}=(0 \quad 0)^T$ is a zero matrix of appropriate dimension.
Denoting the $n\times n$ identity matrix as $I_n$ we thus have $$M_2=\begin{pmatrix} 1&0& a_1 \\ 0&1&a_2 \\ 0&0&B\end{pmatrix}=\begin{pmatrix} I_2 &a_{2\times 1}\\ 0_{1\times 2} & B \end{pmatrix}\\ \implies MM_2=\begin{pmatrix} A_{2\times 2} & a_{2\times 1} \\ b_{1\times 2} & B\end{pmatrix}\begin{pmatrix} I_2 &a_{2\times 1}\\ 0_{1\times 2} & B \end{pmatrix}\\ = \begin{pmatrix} A_{2\times 2}I_2 + a_{2\times 1}0_{1\times 2} &A_{2\times 2}a_{2\times 1}+a_{2\times 1}B\\ b_{1\times 2}I_2+B0_{1\times 2} & b_{1\times 2}a_{2\times 1}+B^2 \end{pmatrix}$$
Note that
$(1)$ any $n\times m$ matrix multiplied with a $m\times p$ zero matrix will give a $n\times p$ zero matrix.
$(2)$ any $n\times m$ matrix multiplied with the identity matrix $I_m$ will give itself as the product.
Thus $$A_{2\times 2}I_2 + a_{2\times 1}0_{1\times 2}=A_{2\times 2}\\ b_{1\times 2}I_2+B0_{1\times 2}=b_{1\times 2} \\ \implies MM_2=\begin{pmatrix} A_{2\times 2} & A_{2\times 2}a_{2\times 1}+a_{2\times 1}B\\ b_{1\times 2} & b_{1\times 2}a_{2\times 1}+B^2\end{pmatrix}$$
Note that the first column of $MM_2$ in the partitioned form already matches the first column of the partitioned form of $M_1$ from above. I'll leave the rest of it to you, where you can try to use the fact that $M$ is orthogonal, i.e. $MM^T=M^TM=I_3$. You can find $M^T$ given the partitioned form of $M$ like this.
N.B. The problem as stated is probably wrong, because the last (second) column of partitioned form of $M_1$ is the last column of $MM^T$, but the last (second) column of $MM_2$ is the last column of $M^2$, so a valid counter-example would be any non-symmetric orthogonal matrix $M$.
