Note: This is essentially the same as Justin's answer, only expanded and explained with more details. You'll recognize my final equations as the same as Justin's final equations, after suitable renaming of the variables.
The dual basis to \beta' consists of functionals $\mathbf{f}_1$ and $\mathbf{f}_2$ such that $\mathbf{f}_1(x_1+2x_2) = 1$ and $\mathbf{f}_1(3x_1+4x_2) = 0$; and $\mathbf{f}_2(x_1+2x_2) = 0$ and $\mathbf{f}_2(3x_1+4x_2) = 1$.
You already have functionals $\phi_1$ and $\phi_2$ satisfying $\phi_i(x_j) = \delta_{ij}$ (Kronecker's delta). And you know that you can write $\mathbf{f}_1$ and $\mathbf{f}_2$ in terms of $\phi_1$ and $\phi_2$ (because the latter form a basis). So you know you can find scalars $\alpha_1,\beta_1$, $\alpha_2,\beta_2$ such that $\begin{align*} \mathbf{f}_1 &= \alpha_1\phi_1 + \beta_1\phi_2\\ \mathbf{f}_2 &= \alpha_2\phi_1 + \beta_2\phi_2. \end{align*}$ What is the value of $\alpha_1\phi_1+\beta_1\phi_2$ on $3x_1+4x_2$? Well, $\begin{align*} (\alpha_1\phi_1+\beta_1\phi_2)(3x_1+4x_2) &= \alpha_1\phi_1(3x_1+4x_2) + \beta_1\phi_2(3x_1+4x_2)\\ &= \alpha_1\Bigl( 3\phi_1(x_1) + 4\phi_1(x_2)\Bigr) + \beta_1\Bigl(3\phi_2(x_1) + 4\phi_2(x_2)\Bigr)\\ &= \alpha_1\Bigl(3(1) + 4(0)\Bigr) + \beta_1\Bigl( 3(0) + 4(1)\Bigr)\\ &= 3\alpha_1 + 4\beta_1. \end{align*}$ Similarly, $\begin{align*} (\alpha_2\phi_1 + \beta_2\phi_2)(3x_1+4x_2) &=3\alpha_2 + 4\beta_2\\ (\alpha_1\phi_1+\beta_1\phi_2)(x_1+2x_2) &= \alpha_1+2\beta_1\\ (\alpha_2\phi_1+\beta_2\phi_2)(x_1+2x_2) &= \alpha_2+2\beta_2. \end{align*}$ But you know what values you want; you want: $\begin{align*} 1 = \mathbf{f}_1(x_1+2x_2) &= (\alpha_1\phi_1 + \beta_1\phi_2)(x_1+2x_2)\\ &= \alpha_1 + 2\beta_1;\\ 0 = \mathbf{f}_1(3x_1+4x_2) &= (\alpha_1\phi_1 + \beta_1\phi_2)(3x_1+4x_2)\\ &= 3\alpha_1 + 4\beta_1;\\ 0 = \mathbf{f}_2(x_1+2x_2) &= (\alpha_2\phi_1 + \beta_2\phi_2)(x_1+2x_2)\\ &= \alpha_2 + 2\beta_2;\\ 1 = \mathbf{f}_2(3x_1+4x_2) &= (\alpha_2\phi_1+\beta_2\phi_2)(3x_1+4x_2)\\ &= 3\alpha_2 + 4\beta_2. \end{align*}$ So now you have a system of four equations in four unknowns: $\begin{array}{rcccccccl} \alpha_1 & + & 2\beta_1 & & & & &=& 1\\ 3\alpha_1 & + & 4\beta_1 && & & & = &0\\ & & & & \alpha_2 & + & 2\beta_2 &=& 0\\ & & & & 3\alpha_2 &+& 4\beta_2 &=& 1. \end{array}$ Solving it will give you the coefficients that express the dual basis of \beta' in terms of $\phi_1$ and $\phi_2$.