I need to double check if my calculations were correct for just these 2 probabilities: $$P(M=m\mid F=f) = \frac{144+36+192+448}{10000} = 0.082$$ $$P(G=g\mid F=f) = \frac{144+1008+36+252}{10000} = 0.144$$ Not very sure if I got the denominator correct, whether I should consider total probability values or just the values concerning the attributes involved. Thanks.
On
No, your calculations are wrong.
For an estimate of $P(M=m\mid F=f)$ you have to count all the cases when $\{F=f\}$ occurred. There are $2800$ such cases; among these cases the number of occurrences of $\{M=m\}$ is $820$. That is
$$P(M=m\mid F=f)\approx\frac{820}{2800}\approx 0.3. $$
For the other probability - the same way
$$P(G=g\mid F=f)\approx\frac{1440}{2800}\approx0.51$$
where $1440$ is the number of occurrences of $\{G=g\}$ when $\{F=f\}$ occurred.
Look to the definition of conditional probability.
$\begin{align}\mathsf P(G=g\mid F=f) ~&=~ \dfrac{\mathsf P(G=g~\cap~ F=f)}{\mathsf P(F=f)} \\[2ex] &=~ \dfrac{144+1008+36+252}{144+1008+36+252~+~192+216+448+504}\cdot\color{silver}{\require{cancel}\dfrac{\cancel{1/10000}}{\cancel{1/10000}}} \\[4ex] \mathsf P(M=m\mid F=f) ~&=~ \dfrac{\mathsf P(M=m~\cap~F=f)}{\mathsf P(F=f)} \\[2ex] &=~ \dfrac{144+36+192+448}{144+1008+36+252~+~192+216+448+504}\end{align}$