a. Assume $\ket{\psi}$ is not entangled with respect to this decomposition. Then (still using the 0/1/2/3 representation for each two-qubit part), $\ket{\psi}$ can be written as
(1)We can see that $a_0 b_0 = \frac{1}{2}$, so $a_0$ is non-zero; and $a_1 b_1 = \frac{1}{2}$, so $b_1$ is non-zero. But we can also see that $a_0 b_1 = 0$. This is impossible. Hence $\ket{\psi}$ must in fact be entangled with respect to this decomposition.
b. Let us write $\ket{\psi}$ as $\frac{1}{2}(\ket{0000} +\ket{0101} +\ket{1010} + \ket{1111})$.
First consider the one qubit / three qubit decomposition where the one qubit is the leftmost in this representation. Suppose $\ket{\psi}$ is not entangled with respect to this decomposition. Then $\ket{\psi}$ can be written as
(2)We can see that $a_0 b_0 = \frac{1}{2}$, so $a_0$ is non-zero; and $a_1 b_2 = \frac{1}{2}$, so $b_2$ is non-zero. But we can also see that $a_0 b_2 = 0$. This is impossible. Hence $\ket{\psi}$ must in fact be entangled with respect to this decomposition.
For each of the other three possible positions for the single qubit, a similar argument can be constructed. Hence $\ket{\psi}$ is entangled with respect to any of these four decompositions.