You know that:
$$\begin{pmatrix}a_1\\a_2\end{pmatrix}\otimes\begin{pmatrix}b_1\\b_2\end{pmatrix}=\begin{pmatrix}a_1b_1\\a_1b_2\\a_2b_1\\a_2b_2\end{pmatrix}=\frac{1}{\sqrt{2}}\begin{pmatrix}\frac12+\frac{\mathrm{i}}{4}\\\frac12+\mathrm{i}\sqrt{\frac{7}{16}}\\\frac12+\frac{\mathrm{i}}{4}\\\frac12+\mathrm{i}\sqrt{\frac{7}{16}}\end{pmatrix}$$
Assuming you don't see any obvious structure (which isn't the case here, so we could speed up the computation a bit in this case), you can proceed using the fact that:
$$\left|a_1b_1\right|^2+\left|a_1b_2\right|^2=\left|a_1\right|^2$$
Using similar equations, you can find the squared modulus of each coefficient. You then want to find their argument. Let us the arguments of the $a_i$ as $\theta_i$ and those of $b_i$ as $\varphi_1$. The argument of $a_ib_j$ is then $\theta_i+\varphi_j\pmod{2\pi}$. You can thus deduce each phase from the equations.
Applying this to this example, we find:
$$\left|a_1\right|^2=\frac12\left|\frac12+\frac{\mathrm{i}}{4}\right|^2+\frac12\left|\frac12+\mathrm{i}\sqrt{\frac{7}{16}}\right|^2=\frac{5}{32}+\frac{11}{32}=\frac12$$
Computing $\left|a_2\right|^2$ leads to the exact same computation. We also have:
$$\left|b_1\right|^2=\frac12\left|\frac12+\frac{\mathrm{i}}{4}\right|^2+\frac12\left|\frac12+\frac{\mathrm{i}}{4}\right|^2=\frac{5}{16}$$
We could perform a similar computation for $\left|b_2\right|^2$, or simply use the fact that $\left|b_1\right|^2+\left|b_2\right|^2=1$ to find $\left|b_2\right|^2=\frac{11}{16}$. We now want to compute the associated phases. We have:
$$\begin{align}
\theta_1+\varphi_1&=\arctan\left(\frac12\right)\\
\theta_1+\varphi_2&=\arctan\left(\frac{\sqrt{7}}{2}\right)\\
\theta_2+\varphi_1&=\arctan\left(\frac12\right)\\
\theta_2+\varphi_2&=\arctan\left(\frac{\sqrt{7}}{2}\right)
\end{align}$$
We thus find $\theta_1=\theta_2$, which simplifies the equations to:
$$\begin{align}
\theta_1+\varphi_1&=\arctan\left(\frac12\right)\\
\theta_1+\varphi_2&=\arctan\left(\frac{\sqrt{7}}{2}\right)\end{align}$$
Note that we won't have enough equations to fully determine all the variables. This is expected: note that if you apply a global phase of $\theta_1$ on the first qubit and a global phase of $-\theta_1$ on the second one, you will end up with the same vector. We can thus consider $\theta_1$ as a variable and express $\varphi_1$ and $\varphi_2$ as functions of $\theta_1$, which finally fives us all of our solutions:
$$a_1=\frac{1}{\sqrt{2}}\mathrm{e}^{\mathrm{i}\theta_1}$$
$$a_2=\frac{1}{\sqrt{2}}\mathrm{e}^{\mathrm{i}\theta_1}$$
$$b_1=\frac{\sqrt{5}}{4}\mathrm{e}^{\mathrm{i}\left(\arctan\left(\frac12\right)-\theta_1\right)}$$
$$b_2=\frac{\sqrt{11}}{4}\mathrm{e}^{\mathrm{i}\left(\arctan\left(\frac{\sqrt{7}}{2}\right)-\theta_1\right)}$$
In particular, for $\theta_1=0$, we have:
$$a_1=\frac{1}{\sqrt{2}}$$
$$a_2=\frac{1}{\sqrt{2}}$$
$$b_1=\frac12+\frac{\mathrm{i}}{4}$$
$$b_2=\frac12+\mathrm{i}\frac{\sqrt{7}}{4}$$