Writing (h+t)(x) is equivalent to writing h(x)+t(x), same goes for (h ⋅ t)(x), as for h[t(x)], it's like replacing what should be x with t(x), so instead of writing x in 2x-5 we write the whole t(x) value so it becomes 2(6x+4)-5
Part A:
(h+t)(x) = h(x) + t(x) = (2x-5) + (6x+4) = 2x+6x + (-5)+4 = 8x - 1
Part B:
(h.t)(x) = h(x) . t(x) = (2x-5) . (6x+4) = (2x).(6x) + (2x) . 4 + (-5).(6x) + (-5).4
= 2.6.x.x + 2.4.x + (-5).6.x + (-5).4
= 12x^2 + 8x - 30x - 20
= 12x^2 + (8 - 30)x -20
= 12x^2 - 22x - 20
Part C:
h[t(x)] = 2(6x+4) -5 = 2.6x +2.4 - 5
= 12x + 8 - 5
= 12x +3
I hope that was clear.