Given that I know $$D\left(P\parallel Q\right)<\alpha,$$ can I say anything about $D\left(Q\parallel P\right)$ in terms of an upper bound on it?


Also, given this upper bound on $D\left(P\parallel Q\right)$, can I deduce some upper bound on $\left|P\left(A\right)-Q\left(A\right)\right|$ for some arbitrary event $A$?



Relation between the two possible KL divergences of two distributions
The answer to your first question is no: $D(Q||P)$ may be however large while $D(P||Q)$ is however small. E.g., let $P$ have masses $s$ and $1-s$ at points $0$ and $1$, respectively, and let $Q$ have masses $t$ and $1-t$ at points $0$ and $1$, respectively, where $0<s,t<1$. Then 
\begin{equation}
 D(P||Q)=s\ln\frac st+(1-s)\ln\frac{1-s}{1-t},
\end{equation}
\begin{equation}
 D(Q||P)=t\ln\frac ts+(1-t)\ln\frac{1-t}{1-s}. 
\end{equation}
Let now $t\downarrow0$ and $s=te^{-1/t^2}$. Then $D(P||Q)\to0$ whereas $D(Q||P)\to\infty$. 


The answer to your second question is yes: [Pinsker's inequality](https://en.wikipedia.org/wiki/Pinsker%27s_inequality) states that 
\begin{equation}
 \sup\_A|P(A)-Q(A)|\le\sqrt{\tfrac12\,D(P||Q)}. 
\end{equation}


