Probability for Type I and Type II error 
P$\bigg(20 < \bar{x} < 35 | \bar{x} \sim N(32, \frac{25^2}{30})\bigg)$

I know the answer is $18.7\%$ (according to my notes) I am not sure how to get to this value. 
Also, how will the method change if I had

P$\bigg(\bar{x} <20 \bigcup \bar{x} > 35 | \bar{x} \sim N(30, \frac{25^2}{30})\bigg)$

For this, the answer is $15\%$.

I am getting different number for your first expression. But this does not matter. (Also, note that your notation $\mathbb{P}[\bar{x}<20|\bar{x}\sim\mathcal{N}(\mu,\sigma)]$ is somewhat nonstandard since it is used for 'conditional on' but you are conditioning on $\bar{x}$ having some distribution, which is not an event. What I think you mean is $\mathbb{P}[\bar{x}<20]$ knowing that $\bar{x}\sim\mathcal{N}(\mu,\sigma)$.)
In the first expression you are calculating $\Phi_{\mu,\sigma}(35)-\Phi_{\mu,\sigma}(20)$, where $\Phi_{\mu,\sigma}$ is cdf of normal distribution with mean $\mu$ and standard deviation $\sigma$ (so that in your case $\mu=32$ and $\sigma=\frac{25}{\sqrt{30}}$). You can use Excel or online calculator to calculate the number.
Alternatively, you can note a well known fact that if $X\sim\mathcal{N}(\mu,\sigma)$, then $\frac{X-\mu}{\sigma}\sim\mathcal{N}(0,1)$. In other words, you know $\bar{x}\sim\mathcal{N}(32,\frac{25}{\sqrt{30}})$ and $$\mathbb{P}[20<\bar{x}<35]=\mathbb{P}\left[\frac{20-32}{\frac{25}{\sqrt{30}}}<\frac{\bar{x}-32}{\frac{25}{\sqrt{30}}}<\frac{35-32}{\frac{25}{\sqrt{30}}}\right]$$ where $\frac{\bar{x}-32}{\frac{25}{\sqrt{30}}}\sim\mathcal{N}(0,1)$. In this case you do not need to calculate cdf of $\mathcal{N}(\mu,\sigma)$ but of $\mathcal{N}(0,1)$.
For the second quantity, you are looking for $\Phi_{\mu,\sigma}(20)=\mathbb{P}[\bar{x}<20]$ and $1-\Phi_{\mu,\sigma}(35)=\mathbb{P}[\bar{x}>35]$.
