This is a mathematical statistics question, and I'm trying to make sure I understand how to compute all of the parts.
A machine stamps out automobile parts. When working correctly, $\mu=1.6$ and $\sigma=0.22$ pounds. To test if the machine is working properly, the staff samples $40$ parts and weigh them. They are testing the following hypothesis $H_0:\mu=1.6$ vs $H_1:\mu>1.6$. They will reject the part if $\bar{y}\geq1.67$ We want to compute both type i and type ii errors.
To make sure I understand this correctly.
$\alpha=P(\text{Reject $H_0$}|\text{H_0 is true})=P\left(z>\frac{1.67-1.6}{.22/\sqrt{40}}\right)=.0222$
Suppose that $\mu=1.68$
$\beta=P(\text{Fail to reject $H_0$}|\text{$H_1$ is true})=P(\bar{y}<1.67|\mu=1.68)=P\left(\frac{1.67-1.68}{.22/\sqrt{40}}\right)=.3859$
Your work is correct, although not entirely precise to the stated number of digits, due to roundoff error or approximation error. To four decimal places, the probabilities are $$\alpha = 0.0221, \quad \beta = 0.3869.$$ But your methodology is correct.
One more thing: You should write for the calculation of $\beta$ $$\Pr[\bar y < 1.67 \mid \mu = 1.68] = \Pr\left[Z < \frac{1.67 - 1.68}{0.22/\sqrt{40}}\right] \approx 0.3869.$$ You omitted the random variable $Z$ in the probability; you correctly included it in your calculation for $\alpha$.
For a verbal interpretation of $\alpha$ and $\beta$, what the answer tells us is that the rejection criterion that is selected, $\bar y \ge 1.67$, indicates that the staff is willing to accept only a $2.21\%$ chance of incorrectly concluding that the machine is producing parts that are too heavy, at the expense of the possibility of failing to detect that the parts are in fact too heavy $38.69\%$ of the time.