Prove that the circle which contains ATB and incircle of ABC touch in one point(T). Incircle of $ABC$ touches $AC$ in $D$, $BC$ in $E$ and $AB$ in $K$. $J$ is the center of the excircle which touches the side $AB$. The circumcircle of $ADJ$ and $BEJ$ intersect in point $J$ and $T$. Prove that the circumcircle of $ATB$ and incircle of $ABC$ touch in one point.
I got that $T$ has to be on incircle of $ABC$. It also looks like that $K$ is on $JT$. One thing which could help is that $T$, center of the incircle of $ABC$ and center of the circumcircle of $ATB$ (labeled $M$ below) are colinear.

 A: First, observe that $\angle DTE = \angle DTJ + \angle JTE = \left(\pi - \angle JAD\right) + \left(\pi - \angle EBJ\right) = 2\pi - \angle JAD - \angle EBJ = 2\pi - \left(\frac \pi 2 + \frac \alpha 2\right) - \left(\frac \pi 2 + \frac \beta 2\right) = \pi - \frac \alpha 2 - \frac \beta 2.$
On the other hand, $\angle EKD = \pi - \angle DKA - \angle BKE = \pi - \left(\frac \pi 2 - \frac \alpha 2\right) - \left(\frac \pi 2 - \frac \beta 2\right) = \frac \alpha 2 + \frac \beta 2$.
Thus $\angle DTE + \angle EKD = \pi$ which implies the fact you mentioned: $T$ lies on the incircle $o$ of triangle $ABC$.
Note that $\angle ATJ = \angle ADJ = \pi - \angle JDC = \pi - \angle CEJ = \angle JEB = \angle JTB$, so $TJ$ is the angle bisector of angle $ATB$.
Notice that $DK \perp AI \perp AJ$, so $DK \parallel AJ$. Moreover, $\angle TJA = \pi - \angle ADT = \angle TDC = \angle TKD$. Therefore $TK \parallel TJ$, so $K \in TJ$. Since $TJ$ is the angle bisector of angle $ATB$, we deduce that $\angle ATK = \angle KTB$.
Let $TK$ intersect circumcircle $\omega$ of triangle $ATB$ at points $T$ and $X$. Then $K$ is the midpoint of arc $AB$ of $\omega$ because $TK$ is bisector of angle $ATB$. Let $l$ be the tangent line to $\omega$ at $X$. Then $AB \parallel l$. 
Consider homothety centered at $T$ which maps $X$ to $K$. Then $l$ is mapped to $AB$ so $\omega$ is mapped to a circle passing through $T$ which is tangent to line $AB$ at point $K$ - this circle is $o$. As a consequence, $\omega$ and $o$ are tangent to each other at point $T$. 
A: Since no one has come up with an answer yet, here is a rather brute-force computation on homogeneous coordinates to verify this fact. Without loss of generality I'll choose my coordinate system in such a way that the incircle becomes the unit circle, with $K=(1:0:1)$ at the $0°$ position. Then $D$ and $E$ can be described by one peal parameter each, using the half-angle formula. This gives us
\begin{align*}
E&=(t^2-1:2t:t^2+1) & D&=(u^2-1:2u:u^2+1)
\end{align*}
Everything else can be expressed in terms of $t$ and $u$. The unit circle has the diagonal matrix $U=\operatorname{diag}(1,1,-1)$, and multiplying that matrix with the vecotr of a point gives the tangent in that point. Computing the cross product between two lines gives their point of intersection. So you get
\begin{align*}
A&=(U\cdot D)\times(U\cdot K)=(u:1:u) \\
B&=(U\cdot E)\times(U\cdot K)=(t:1:t) \\
C&=(U\cdot D)\times(U\cdot E)=(tu - 1 : t + u : tu + 1)
\end{align*}
Next, we need $J$. I'd construct that by intersecting the line orthogonal to $AI$ through $A$ with the one orthogonal to $BI$ through $B$. To form a line connecting two points, you again compute the cross product. If you set the last coordinate of the resulting vector to $0$ you obtain a vector which describes a point infinitely far away in a direction orthogonal to the line. Connecting that with another point gives an orthogonal line. Dropping the last coordinate can be formulated by multiplication with the matrix $F=\operatorname{diag}(1,1,0)$.
\begin{align*}
J &= (A\times(F\cdot(A\times I)))\times(B\times(F\cdot(B\times I)))
= (tu - 1 : t + u : tu)
\end{align*}
Next we need circles. I computed circles as conics through the ideal circle points $Q=(1:i:0)$ and $\bar Q=(1:-i:0)$. (Usually I'd call these points $I$ and $J$, but those letters are already taken in your problem statement.) You can find the matrix for the circle through three points $a,b,c$ by computing
\begin{align*}
M_1 &= \det(a,c,Q)\cdot \det(b,c,\bar Q)\cdot (b\times Q)\cdot(a\times\bar Q)^T \\
M_2 &= (M_1 - \bar M_1)+(M_1 - \bar M_1)^T
\end{align*}
Using this approach, your circles are described by
\begin{align*}
\bigcirc_{ADJ}=\begin{pmatrix}
2 t u & 0 & 1 - t u -  u^{2} \\
0 & 2 t u & - t - 2 u \\
1 - t u -  u^{2} & - t - 2 u & 2 u^{2} + 2
\end{pmatrix}
\\
\bigcirc_{BEJ}=\begin{pmatrix}
2 t u & 0 & 1 - t^{2} -  t u \\
0 & 2 t u & -2 t -  u \\
1 - t^{2} -  t u & -2 t -  u & 2 t^{2} + 2
\end{pmatrix}
\end{align*}
Subtracting these matrices, you obtain a degenerate conic which factors into the line at infinity and the line connecting the two points of intersection. That line connecting the intersections is $g=(t + u : 1 : -t - u)$. It's point at infinity is $G=(-1:t+u:0)$. This allows you to find the point $T$ as the intersection of that line with one of the two circles:
\begin{align*}
T&=G^T\cdot\bigcirc_{ADJ}\cdot G\cdot J - 2\cdot J^T\cdot\bigcirc_{ADJ}\cdot G\cdot G\\
T&=%(t^2 + 2tu + u^2 - 1, 2t + 2u, t^2 + 2tu + u^2 + 1)
((t+u)^2-1 : 2(t+u) : (t+u)^2+1)
\end{align*}
One can already recognize the half angle format of this point, so it will lie on the unit circle. You need one last circle, $ABT$:
\begin{align*}
\bigcirc_{ABT}=\begin{pmatrix}
4 t u & 0 & 1 - t^{2} - 2 t u -  u^{2} \\
0 & 4 t u & -2 t - 2 u \\
1 - t^{2} - 2 t u -  u^{2} & -2 t - 2 u & 2 t^{2} + 2 u^{2} + 2
\end{pmatrix}
\end{align*}
Now you can verify that $T^T\cdot U\cdot T=T^T\cdot\bigcirc_{ABT}\cdot T=0$, so the point $T$ lies on both these circles. Furthermore, the tangent to the circle is the same in both cases as well, up to a scalar factor. So we have a single touching point.
\begin{align*}
U\cdot T \sim \bigcirc_{ABT}\cdot T &=
(t^2 + 2tu + u^2 - 1 : 2t + 2u : - t^2 - 2tu - u^2 - 1)
\end{align*}
All through this post, I've cleared common denominators from the homogeneous coordinates and circle matrices to keep things simple. And I'm happy that I could write all of this without needing a single square root. I think about my homogeneous coordinates as column vectors, even though I wrote them as rows to save space. This explains which expressions I transposed in some computations and which I left as columns.
Both your assumptions are true, by the way. $K$ lies on $J\times T=(t+u:1:-t-u)$ and the center of the circle $ABT$, which is the point $M=((t+u)^2 - 1 : 2(t+u) : 4tu)$, lies on the line $T\times I=(-2(t+u) : (t+u)^2 - 1 : 0)$. $M$ does not lie on the incircle, even though it does look that way in the figure.
Seeing that the half angle tangent parameter of $T$ is $t+u$ tells me that the line $DE$ will intersect $AB$ in a point and that point connected to $T$ will intersect the incircle in another point, which also lies on $KI$. Not sure whether this is of any use to anybody.
