Note that $FL_{.75}^c$ (a full hour is used and the student is not ready in less than $.75$ hours) is exactly the same event as $F$ (a full hour is used). You could say that $F\subset L_{.75}^c$ and consequently $F\cap L_{.75}^c=F$.
$P(L_x^c)=1-\frac{x}{2}$ leads directly to $P(L_{0.75}^c)=1-\frac{0.75}{2}=0.625$
The setup is incorrect. You appear to have the conditional probabilities for the events (alarm given cooking and smoke), rather than the joint probabilities.
You want. $\mathsf P(A{=}\mathsf T) ~=~ {{\mathsf P(S{=}\mathsf T,C{=}\mathsf T)~\mathsf P(A{=}\mathsf T\mid S{=}\mathsf T,C{=}\mathsf T)}+{\mathsf P(S{=}\mathsf T,C{=}\mathsf F)~\mathsf P(A{=}\mathsf T\mid S{=}\mathsf T,C{=}\mathsf F)}+{\mathsf P(S{=}\mathsf F,C{=}\mathsf T)~\mathsf P(A{=}\mathsf T\mid S{=}\mathsf F,C{=}\mathsf T)}+{\mathsf P(S{=}\mathsf F,C{=}\mathsf F)~\mathsf P(A{=}\mathsf T\mid S{=}\mathsf F,C{=}\mathsf F)}}$
You have, $\mathsf P(A{=}\mathsf T\mid S{=}\mathsf T,C{=}\mathsf T)=0.45, \mathsf P(A{=}\mathsf T\mid S{=}\mathsf T,C{=}\mathsf F)=0.15,$ et. cetera.
You are missing; $\mathsf P(S{=}\mathsf T,C{=}\mathsf T), \mathsf P(S{=}\mathsf T,C{=}\mathsf F), \mathsf P(S{=}\mathsf F, C{=}\mathsf T), \mathsf P(S{=}\mathsf F,C{=}\mathsf F)$
With the added information, you have $\mathsf P(S{=}\mathsf T)=0.27, \mathsf P(C{=}\mathsf T)=0.42$ and from the diagram, they are independent random variables, so $\mathsf P(S{=}\mathsf T, C{=}\mathsf T)=\mathsf P(S{=}\mathsf T)~\mathsf P(C{=}\mathsf T) = (0.27)(0.42) \\ \mathsf P(S{=}\mathsf T, C{=}\mathsf F)=\mathsf P(S{=}\mathsf T)~\mathsf P(C{=}\mathsf F) = (0.27)(0.58)\\\text{etc.} $
Then you now have enough information to calculate the join probability masses: $$\mathsf P(A{=}\mathsf T,S{=}\mathsf T,C{=}\mathsf T)=(0.27)(0.42)(0.45)=0.05103$$
And so forth.
Best Answer
$\checkmark$ Yes.
Each entry is something like: $$\mathsf P(A\cap \neg B\cap A_B\cap \neg B_A)= \mathsf P( A)\,\mathsf P(\neg B)\,\mathsf P(A_B\mid A, \neg B)\,\mathsf P(\neg B_A\mid A, \neg B)$$
The rule is the product rule for conditional probabilities.
For any events $X,Y$ then $\mathsf P(X\cap Y)=\mathsf P(X)\mathsf P(Y\mid X)$, and if $X$ and $Y$ are independent then also $\mathsf P(Y\mid X)=\mathsf P(Y)$.
When you have the table: $$\mathsf P(B\mid A, A_B,-B_A) = \dfrac{\mathsf P(A\cap B\cap A_B\cap \neg B_A)}{\mathsf P(A\cap B\cap A_B\cap \neg B_A)+\mathsf P(A\cap \neg B\cap A_B\cap \neg B_A)}$$
Using the Product Rule and the Law of Total Probability.