The number of values of k for which the system of linear equations, (2k+1)x + 5ky = k+2 and kx + (k+2)y = 2 has no solution, is:

A non-homogeneous system of linear equations has a unique non-trivial solution if and only if its determinant is non-zero. If this determinant is zero, then the system has either no nontrivial solutions or an infinite number of solutions.

The given system of equations is:
(2k+1)x + 5ky = k+2
kx + (k+2)y = 2

For the system to have no solution, the determinant of the coefficient matrix must be zero, and at least one of the determinants of the augmented matrix must be non-zero. Let's find the determinant of the coefficient matrix:

| 2k+1 5k |
| k k+2 |

= (2k+1)(k+2) - 5k² = 2k² + 5k + 2 - 5k² = -3k² + 5k + 2

Setting this determinant to zero:
-3k² + 5k + 2 = 0
3k² - 5k - 2 = 0
(3k + 1)(k - 2) = 0
k = 2 or k = -1/3

Now let's check the augmented matrix determinants. If k = 2:

| 5 10 | | 4 | | 5 4 | | 10 4 |
| 2 4 | = | 2 | | 2 2 | = | 4 2 |

5(4) - 10(2) = 0
5(2) - 4(2) = 2 ≠ 0
Therefore, when k = 2 there is no solution.

If k = -1/3:

| 1/3 -5/3 | | 5/3 | | 1/3 5/3 |
| -1/3 5/3 | = | 2 |

(1/3)(5/3) - (-5/3)(-1/3) = 0
(1/3)(2) - (5/3)(-1/3) = 7/9 ≠ 0
Therefore, when k = -1/3 there is no solution.

Thus, there are 2 values of k for which the system has no solution.