P(fish for $>2$ hours) = $P(k=0,\tau =2)$ = $e^{-0.6\times 2}$

P(fish for $>2$ but $<5$ hours) = P(first catch in $[2,5]$ hours) = $P(k=0,\tau =2)(1-P(k=0,\tau =3)$ which is no fish in $[0,2]$ but at least $1$ fish in the next $3$ hours (which will be independent of first $2$ hours)

P(catch at least two fish) = P(at least $2$ catches before $2$ hours) = $1-P(k=0,\tau =2)-P(k=1,\tau =2)$

$E[fish]$ has two possibilities, either single fish after $2$ hours, or many fist before $2$ hours. $$\begin{array}{rl}E[fish]& =E[fish|\tau \le 2](1-P(\tau >2))+E[fish|\tau >2]P(\tau >2)\\ & =(0.6\times 2)\times (1-P(k=0,\tau =2))+1\times P(k=0,\tau =2)\end{array}$$

$E[$Total fishing time$]$ = $2+P(k=0,\tau =2)\frac{1}{\lambda }$, since we fish for atlest $2$ hours

$E[$future fishing time $|$ fished for two hours$]$ can be obtained using the memoryless property of Poisson process. The expected time till first arrival is independent of what has happened till now. Thus, $E[T_1]=\frac{1}{\lambda }$