On your first question: no, because if $P(A)<1$ then conditional independence wrt $A$ does not imply independence. See the answer of littleO.
Further $P(B\mid C)$ cannot be expressed in $P(A)$, $P(B\mid A)$ and $P(C\mid A)$.
This is illustrated by as follows.
If $B=A$ and $C=\Omega$ then:$$P(B\cap C\mid A)=1=P(B\mid A)P(C\mid A)$$
and this with $P(B\mid A)=P(C\mid A)=1$
But if $B=\Omega$ and $C=A$ then this is also true.
In the first case $P(B\mid C)=P(A)<1$ but in the second case $P(B\mid C)=1$.
This in spite of the fact that in both cases $P(A)$, $P(B\mid A)$ and $P(C\mid A)$ have the same values.
edit for completeness.
There is an expression for $P(A)$, $P(B\mid A)$, $P(C\mid A)$ and $P(C)$.
We have: $$P(B\cap C)=P(B\cap C\mid A)P(A)=P(B\mid A)P(C\mid A)P(A)$$ so that: $$P(B\mid C)=\frac{P(B\cap C)}{P(C)}=\frac{P(B\mid A)P(C\mid A)P(A)}{P(C)}$$
Further note that: $$\frac{P(C\mid A)P(A)}{P(C)}=\frac{P(C\cap A)}{P(C)}=P(A\mid C)$$
So we can also write:$$P(B\mid C)=P(B\mid A)P(A\mid C)$$