Qubit measurement of the state $frac{1}{sqrt2}|00rangle+frac{i}{2}|01rangle-frac{1}{2}|11rangle$

If first qubit is $|1rangle$ there is no other possibility than second qubit to be $|1rangle$ as well since probability of state $|10rangle$ is zero. Hence probability of measuring $|1rangle$ in second qubit is $1$.

In case first qubit is $|0rangle$ there are two possibe results: $|00rangle$ or $|01rangle$. Since probability of state $|00rangle$ is $frac{1}{2}$ and probability of state $|01rangle$ is $frac{1}{4}$, conditional probabilities that second qubit is $|0rangle$ is $frac{2}{3}$ and that it is $|1rangle$ is $frac{1}{3}$.

This is about probabilities of measuring $|0rangle$ or $|1rangle$ in computational basis, regarding quantum state of second qubit before its measurement, please refer to Shai Dashe comment:

Hint: note that $|varphirangle = |0rangle otimes big(frac{1}{sqrt{2}}|0rangle + frac{i}{2}|1ranglebig) + |1rangleotimes big(-frac{1}{2}|1ranglebig)$