If the point $$(1, a)$$ lies between the straight lines $$x + y = 1$$ and $$2(x+y) = 3$$ then $$a$$ lies in interval

The two given lines are
  $$x + y = 1 \; \; -(1)$$
  $$2(x+y) = 3 \;\Longrightarrow\; x + y = \frac32 \; \; -(2)$$

Write each line in the form $$L=0$$:
  $$L_1 : x + y - 1 = 0$$
  $$L_2 : x + y - \frac32 = 0$$

A point $$P(x_0,y_0)$$ lies between two non-parallel lines $$L_1=0$$ and $$L_2=0$$ if the algebraic distances of the point from the lines have opposite signs, i.e.
  $$L_1(x_0,y_0)\;L_2(x_0,y_0) \lt 0.$$ This is because the product is negative only when the point is on opposite sides of the two lines.

For the point $$P(1,a)$$:
  $$L_1(1,a)=1+a-1=a$$
  $$L_2(1,a)=1+a-\frac32=a-\frac12$$

Condition for lying between the lines:
  $$a\left(a-\frac12\right) \lt 0$$

The product of two real numbers is negative when the numbers have opposite signs. Thus
  $$a \gt 0 \quad\text{and}\quad a-\frac12 \lt 0 \quad\Longrightarrow\quad 0 \lt a \lt \frac12$$

Hence $$a$$ must lie in the open interval $$\left(0,\frac12\right).$$

Option D which is: $$\left(0, \dfrac{1}{2}\right)$$