Attempt:
Notice we can write $T = \max (T_1, T_2)$. One can find the pmf:
$$ P(T \leq t) = P(max(T_1,T_2) \leq t) = P(T_1 \leq t, T_2 \leq t) = (1-e^{-t/2})^2$$
therefore $f_T(t) = (1-e^{t/2}) e^{-t/2} = e^{-t/2} - e^{-t}$
Now, $f_{T|T_1=1} (t|t_1) = \dfrac{f_T(t)}{f_{T_1}(1)} = \dfrac{e^{-t/2} - e^{-t}}{e^{-1}} $
Thus, the expectation is
$$ E(T | T_1=1) = \int t \dfrac{e^{-t/2} - e^{-t}}{2e^{-1/2}} dt = e (2 - 1) = \boxed{2e^{1/2}}$$
Now, for $b$, we evaluate $\int_1^{\infty} f_{T_1} (t) = 1 - P(1) = e^{-1/2} $
So, $E(T | T_1 > 1) = e^{1/2} $
Now, Is my approach correct so far? For part c), what would be the best approach? or just brute furce ?
For a) I would consider $$ E(T|T_1 = 1) = E(T|T_2 > 1, T_1 =1) P(T_2 > 1) + E(T|T_2<1, T_1=1)P(T_2<1) $$
The first expected value simplifies to $E(T_2|T_2>1) = 1 +E(T_2)$ using memoryless property and the second one is just 1.
The same approach works for b) $$ E(T|T_1 > 1) = E(T|T_2 > 1, T_1 > 1) P(T_2 > 1) + E(T|T_2<1, T_1> 1)P(T_2<1) $$
Using memoryless property of $T_1$ and $T_2$ the first expected value is $1 + E(T)$ and the second one is $E(T_1|T_1>1)=1+E(T_1)$.
Hope this helps for c) :)