Find $\mathbb{E}[T_1|X(1)=2]$ where $X(t)$ be the Poisson process. 
Let $X(t)$ be the Poisson process with rate $\lambda$. Let $T_n=:\min\{t\geq 0: X(t)=n\}$ be the waiting time for the $n$th signal. Find $\mathbb{E}[T_1|X(1)=2]$.

I am confused about $\mathbb{E}[T_1|X(1)=2]$. Here condition on $X(1)=2$ BUT why we can consider $\min\{t\geq 0: X(t)=1\}$? Is $t\leq 1$?
(Thanks for  @Brian Moehring's comment, I edited my solution.)
I can compute $\mathbb{P}[T_1<s|X(1)=2]$ as following:
$$\mathbb{P}[T_1<s|X(1)=2]=\frac{\mathbb{P}[T_1<s, X(1)=2]}{\mathbb{P}[X(1)=2]}$$
$$=\frac{\mathbb{P}[\text{one event in } [0, s), \text{ one event in  } [s,1]]+\mathbb{P}[\text{two events in } [0, s), \text{ zero event in  } [s,1]]}{\mathbb{P}[X(1)=2]}$$
$$=\frac{\mathbb{P}[\text{one event in } [0, s)]\mathbb{P}[ \text{ one event in  } [s,1]]+\mathbb{P}[\text{two events in } [0, s)]\mathbb{P}[ \text{ zero event in  } [s,1]]}{\mathbb{P}[X(1)=2]}$$
$$=\frac{\lambda se^{-\lambda s} \lambda (1-s)e^{-\lambda (1-s)}+1/2(\lambda s)^2 e^{-\lambda s}e^{-\lambda (1-s)}}{\mathbb{P}[X(1)=2]}=2s(1-s)+s^2=2s-s^2$$

Is it 
  $$\mathbb{E}[T_1|X(1)=2]=\int_0^1 s d(\mathbb{P}[T_1<s|X(1)=2])=\int_0^1 s d(2s(1-s)+s^2)=\int_0^1 s(2-2s)ds?$$

 A: After your correction, your solution looks fine.
As Villa mentioned in the comments, the answer ends up being $1/3$.  We can actually go further than that, however, since we have actually found the distribution of $T_1 \mid (X(1)=2).$  If you look at it briefly, you might notice that $$T_1 \mid (X(1)=2) \stackrel d= \min\{U_1,U_2\}$$
where $U_1,U_2 \stackrel{\text{i.i.d.}}\sim UNIF([0,1]).$
It's not hard to similarly investigate $T_2 \mid (X(1)=2)$ to see it is distributed as $\max\{U_1,U_2\}.$  From here, we can guess [correctly] that $$(T_1, T_2) \mid (X(1)=2) 
 \stackrel d= (U_{(1)}, U_{(2)}),$$
[Here, I adopt the convention $X_{(k)}$ is the $k^\text{th}$ order statistic]

In fact, we can go further in proving for $m,n \geq 0$ the much more general
$$(T_1, T_2, \ldots, T_{m+n}) \mid (X(a)=m) \stackrel d= (U_{(1)}, U_{(2)}, \ldots, U_{(m)}, a+T_1, a+T_2, \ldots, a+T_n)$$
where $U_1, U_2, \ldots, U_m \stackrel{\text{i.i.d.}}\sim UNIF([0,a])$ and $(U_1, \ldots, U_m)$ and $X$ are independent.
