Hi @cucho,
Here’s a fresh derivation, due to my friend Leo Bastos. Please note that the probability he derives and the one I did above are not for the same event. Nevertheless, I believe his derivation is clearer and more useful and hence my post (and edits) above should be all but disregarded.
Let R \in \{0, 1\} be the true disease status and let \gamma_i and \delta_i be the specificity and sensitivity of test i, respectively.
Let Y be the outcome of a re-testing positives only strategy, i.e. Y=1 if tests 1 and 2 are both positive, and Y=0 otherwise. So
\begin{align*}
p_2 := \operatorname{Pr}( Y = 1) & = \operatorname{Pr}(Y = 1 , R = 1) + \operatorname{Pr}(Y = 1 , R = 0),\\
& = \operatorname{Pr}(Y = 1 \mid R = 1) \operatorname{Pr}(R = 1) + \operatorname{Pr}(Y = 1 \mid R = 0) \operatorname{Pr}(R = 0),\\
& = \operatorname{Pr}(T_2 = 1 , T_1 = 1\mid R = 1) \operatorname{Pr}(R = 1) + \operatorname{Pr}(T_2 = 1 , T_1 = 1\mid R = 0) \operatorname{Pr}(R = 0), \\
& = \operatorname{Pr}(T_2 = 1 \mid T_1 = 1, R = 1) \operatorname{Pr}(T_1 = 1\mid R = 1) \operatorname{Pr}(R = 1) + \\
& + \operatorname{Pr}(T_2 = 1 \mid T_1 = 1, R = 0) \operatorname{Pr}(T_1 = 1\mid R = 0) \operatorname{Pr}(R = 0), \\
& = \delta_2 \delta_1 \pi + (1 - \gamma_2)(1 - \gamma_1) (1 - \pi)
\end{align*}
where the last line follows if we assume that the two tests are conditionally independent given R.