Since $|x-7|=3$, the value of $x$ is $3$ units away from $7$. So the two solutions are $x=7+3=10$ and $x=7-3=4$. Their sum is $10+4=14$.

Since $(k,3)$ is on the graph of $y=|x-4|$, substitute $x=k$ and $y=3$ to get $3=|k-4|$. This means $k-4=3$ or $k-4=-3$, so $k=7$ or $k=1$. One possible value is $1$.

Because both expressions equal $y$, set them equal to each other:

Now solve by cases.

If $x\ge 4$, then $|x-4|=x-4$. So

which gives

Then

So

If $x<4$, then $|x-4|=4-x$. So

which simplifies to $4=6$, impossible. Therefore the only solution is $(5,1)$, and the value of $x+y$ is $6$.

For an equation of the form $y=|x-h|+k$, the graph is a V-shape with vertex at $(h,k)$. In the table, the smallest value of $y$ is $4$, and it occurs when $x=3$. Also, the values are symmetric: $y=6$ at $x=1$ and $x=5$, and $y=8$ at $x=-1$ and $x=7$. That means the axis of symmetry is $x=3$, so the vertex is at $(3,4)$. Therefore, $h=3$.

For an equation of the form $|x-4|=k$, the left side is a distance from $x$ to $4$, so it can never be negative. If $k>0$, then there are two solutions: one value of $x$ that is $k$ units to the left of $4$ and one value that is $k$ units to the right of $4$. If $k<0$, there are no real solutions. The only way the equation has exactly one real solution is when the distance is $0$, which happens at $x=4$. Therefore, $k$ must be $0$.