Independent odds, am I (+ friend) seeing this wrong or is there a mistake in the practice exam?

As the other answerers have noted, $0.99 = 1 - 0.1^2$ is indeed the correct answer.

As to what went wrong with the exercise, I suspect the problem statement has a typo, and the correct admission rate should be 80% instead of 90%. That would make option E ($0.96$ $=$ $1 - 0.2^2$) the correct one.

Why do I suspect that? It's because, as you note, for two independent trials with the same success rate, \begin{aligned} {\rm Pr}[\text{one succeeds}] &= 1 - {\rm Pr}[\text{both fail}] \\ &= 1 - (\text{failure rate})^2 \\ &= 1 - (1 - \text{success rate})^2, \end{aligned} which implies that, conversely, $$\text{success rate} = 1 - \sqrt{1 - {\rm Pr}[\text{one succeeds}]}.$$

Applying this "reverse" formula to the given options, we can work out what the admission rate would have to be for each option to be correct:

\begin{aligned} {\rm A}:\ {\rm Pr}[\text{one succeeds}] &= 0.5 &\implies& \text{success rate} = 1 - \sqrt{0.5}\phantom0 \approx 0.292893 \\ {\rm B}:\ {\rm Pr}[\text{one succeeds}] &= 0.65 &\implies& \text{success rate} = 1 - \sqrt{0.35} \approx 0.408392 \\ {\rm C}:\ {\rm Pr}[\text{one succeeds}] &= 0.88 &\implies& \text{success rate} = 1 - \sqrt{0.12} \approx 0.653589 \\ {\rm D}:\ {\rm Pr}[\text{one succeeds}] &= 0.9 &\implies& \text{success rate} = 1 - \sqrt{0.1}\phantom0 \approx 0.683772 \\ {\rm E}:\ {\rm Pr}[\text{one succeeds}] &= 0.96 &\implies& \text{success rate} = 1 - \sqrt{0.04} = 0.8 \end{aligned}

Out of these options, E is the only one where the probability of both students failing to be admitted works out to a nice square ($0.04 = 0.2^2$). For any of the options A to D to be (exactly) correct, the admission rate would have to be a very awkward irrational number.

Of course, it's also possible that the error is in the options themselves, and that the author of the exercise meant to include 0.99 as one of the options. The actual explanation might even be a combination of both possibilities — perhaps whoever wrote the exercise started with an admission rate of 80%, came up with a suitable answer set, and then later decided to change the admission rate to 90% but forgot to update the answers.

Sometimes you have to challenge the options. Your concepts and answer are absolutely correct!

That's the standard way of doing it:

$P(\text{At Least One}) = 1 - P(\text{Both Fail}) = 1 - 0.1 \times 0.1 = 1 - 0.01 = 0.99$

And the not so standard way:

$P(1\text{ in}) + P(2\text{ in Without }1) = 0.9 + (0.9 \times 0.1) = .99$.

Or

$$\begin{align} P(1\text{ in Without }2) + P(2\text{in Without }1) + P(1\text{ in And }2\text{ in}) &= 0.9 \times 0.1 + 0.9 \times 0.1 + 0.9 \times 0.9 \\ &= 0.09 + 0.09 + 0.81 = .99 \end{align}$$

No matter how we cut it, you are right. They are wrong.

(It's probably just a typo.)