Let $H_i$ be the event 'The $i$-th toss comes up heads' and similarly for $T_i$ with tails.
Let $F$ be the event that 'The chosen coin is fair' and $B$ be the event that 'The chosen coin is biased'.
Then
\begin{align}\mathbb P(H_3|H_1\cap H_2)=\frac{\mathbb P(H_3\cap H_1\cap H_2)}{\mathbb P(H_1\cap H_2)}\end{align}
Now, we have
\begin{align}
\mathbb P(H_3\cap H_1\cap H_2)
&=\mathbb P(H_3\cap H_1\cap H_2|F)\cdot \mathbb P(F)+\mathbb P(H_3\cap H_1\cap H_2|B)\cdot \mathbb P(B)\\
&=\frac1{2^3}\cdot\frac12+1\cdot\frac12=\frac1{16}+\frac12=\frac9{16}
\end{align}
and
\begin{align}
\mathbb P(H_1\cap H_2)
&=\mathbb P(H_1\cap H_2|F)\cdot \mathbb P(F)+\mathbb P( H_1\cap H_2|B)\cdot \mathbb P(B)\\
&=\frac1{2^2}\cdot\frac12+1\cdot\frac12=\frac1{8}+\frac12=\frac5{8}=\frac{10}{16}
\end{align}
Hence we have that
\begin{align}\mathbb P(H_3|H_1\cap H_2)
&=\frac{\frac9{16}}{\frac{10}{16}}=\frac9{10},\end{align}
as you claimed.