Number of solutions - please check my solution? 
Determine, in accordance to $k$, how many solutions does the given
  system of equations have: 
  $$ \begin{cases}kx+(k+1)y=k-1\\4x+(k+4)y=k\end{cases} $$ 
  And check, for which values of $k$ this system has exactly one solution lying   within the boundaries of a triangle of vertexes: $A=(0, 0), B=(\frac{2}{3},
 0), C=(0, 2)$.

Can I just solve it using determinants? I mean:
$$D=k(k+4)-4(k+1)=k^2-4=(k+2)(k-2)$$
$$D_x=(k-1)(k+4)-(k+1)k=2k-4$$
$$D_y=k^2-4(k-1)=k^2-4k+4$$
Our system has one solution if $D$ is not $0$. This happens when $k$ is neither $2$ nor $-2$. 
It has no solutions when $D=0$ and ($D_x \neq 0$ or $D_y \neq 0$). $D=0$ for $k=2$ or $k=-2$,  $D_x \neq 0$ when $k \neq 2$ and $D_y \neq 0$ when $k \neq 2$. Though for $k=2$ our $D$ equals $0$, neither $D_x$ nor $D_y$ is $\neq 0$ so it has no solutions only when $k=-2$.
It has infinitely many solutions for $D=0$ (which means $k=2$ or $k=-2$), $D_x = D_y = 0$. This happens when $k=2$.
Summing up: 
$$
\begin{cases}
\text{no solution}, &\text{for } k=-2,
\\
\text{infinitely many solutions}, &\text{for } k=2,
\\ 
\text{one solution}, &\text{for every other } k.
\end{cases}
$$
Is this all right? I need to know that before proceeding to the triangle-thing.

First, let's get back to the mainstream terminology: We have a linear system $Ax=b$ with the data 
$$
\pmatrix{k &k+1\\4 &k+4}\pmatrix{x\\y}=\pmatrix{k-1\\k}
$$
Now if $A$ is invertible then there exists a unique solution which is given by
$$
\pmatrix{x\\y} = \pmatrix{k &k+1\\4 &k+4}^{-1}\pmatrix{k-1\\k} = \frac{1}{k^2-4}\pmatrix{k+4 &-(k+1)\\-4&k}\pmatrix{k-1\\k} = \pmatrix{\frac{2}{k+2}\\ 1-\frac{4}{k+2}}
$$
As you have computed, the easiest way to conclude invertibility is having the determinant nonzero. Computing the determinant gives: $\det A = k^2-4 = (k-2)(k+2)$, Thus, we have a unique solution for all $|k|\neq 2$.
Regarding the zero determinant cases, let's look at the system at those points: 
($k=2$)
$$
\pmatrix{2&3\\4&6}\pmatrix{x\\y} = \pmatrix{1\\2}
$$
We can clearly see that the second equation is the 2-multiple of the first one. Hence we actually only have $2x+3y = 1$
($k=-2$)
$$
\pmatrix{-2&-1\\4&2}\pmatrix{x\\y} = \pmatrix{-3\\-2}
$$
We can see that the right hand side is not in the range of $A$, hence there is no solution. Another way to look at it is rewriting as
$$
\pmatrix{-2\\4}x+\pmatrix{-1\\2}y = \pmatrix{-1\\2}(2x+y) = \pmatrix{-1\\2}\alpha = \pmatrix{-3\\-2}
$$
It is obvious that no alpha can satisfy that equality hence no solution. 
Regarding the triangle question, you can rewrite the constraints as
$$
\pmatrix{-3 &-1\\1 &0\\0&1}\pmatrix{x\\y}\geq \pmatrix{-2\\0\\0}
$$
Now plug in the $k$ dependent solution into this and we get
$$
\pmatrix{1-\frac{2}{(k + 2)}\\
       \frac{2}{(k + 2)}\\
   1 - \frac{4}{(k + 2)}}\geq 0
$$
This gives us the condition on $k$:
$$
0\leq \frac{2}{k+2} \leq \frac{1}{2} \implies k>2
$$
Note that we ruled out $k=2$ case since there are infinitely many solutions in that case violating the requirement given in the question. 
