Question 1.

$2x^2 + kx + 5 = 0$ has no real roots so $D$ < $0$

$k^2 - 40$ < $0$

$(k - \sqrt{40})(k + \sqrt{40})$ < $0$

$k \in (-\sqrt{40}, \sqrt{40})$

$x^2 + (k - 5)x + 1 = 0$ has two distinct real roots so $D$ > $0$

$(k - 5)^2 - 4$ > $0$

$k^2 - 10k + 21$ > $0$

$(k - 3)(k - 7)$ > $0$

$k \in (-\infty, 3) \cup (7, \infty)$

Therefore possible value of k are -6, -5, -4, -3, -2, -1, 0, 1, 2