Question: One about flipping coins. Coin A has 0.6 probability of being heads, Coin B has a 0.3 probability of being heads and Coin C has a 0.1 probability of heads. Given that a coin is flipped and lands on heads, what is the probability that when it is flipped again it will also be heads.
I'm not sure how I can apply Bayes' rule here.

I did a Monte Carole simulation the answer come out to be around 0.46, I am not sure how to give a analytical answer.

Answer: 
(a) Yes this is true, but you cannot think of a $4 \times 4$ matrix with a row of all zeroes as a $3 \times 4$ matrix. But the determinant of any matrix with an all zero row is zero, hence the matrix is not invertible.

(b) Correct

(c) follows straight from the definition. A matrix $A$ is said to be invertible if there exists a matrix $A^{-1}$ such that $AA^{-1} = A^{-1}A = I$. Note by this definition, $A^{-1}$ is also invertible with inverse $A$. To show that $A^2$ is invertible, just note that
$$A^2(A^{-1})^2 = AAA^{-1}A^{-1} = AIA^{-1} = AA^{-1} = I.$$
You can check that $(A^{-1})^2A^2 = I$ as well.
The answer for (b) is good. (If you want to be very precise you'll have to state that you exclude the case of $1 \times 1$ matrices, but often this goes without saying.)

The answer for (a) might be good idea, but not expressed too well. 
A way to argue this is that $A$ invertible, means that $Ax = b$ has a solution whatever the $b$. 
But this is false in this case, as the result of $Ax\_0$ will always yield a vector with $0$ in the row that is the $0$-row in the matrix. 

For (c). Just show that there is a matrix $C$ such that $A^{-1}C=CA^{-1}=I$ and likewise there is a matrix $D$ such that $A^2D = D A^2 = I$. (In the first case take $C = A$ and in the second $D = (A^{-1})^2$.)