$$
Wsintheta = lambda.
$$
Knowing that by increasing $theta$ we also increase $sintheta$ wee see that $lambda$ i.e. the required wavelength also increases so it has to be greater than $511text{ nm}$. If $theta = 15^circ$ we directly calculate

$$
lambda=2.20times10^{-6}text{ m}timessin15^circ = 569text{ nm}
$$
which confirms our prediction.

$$
Wsintheta = lambda
$$
we obtain

$$
W=frac{lambda}{sintheta}=frac{546times10^{-9}text{ m}}{sin16^circ} = 1.98text{ $mu$m}.
$$

$$
Wsintheta = lambda.
$$
The width of the slit here is the width of the door. Also note that from $u=lambdanu$ we obtain $lambda = frac{c}{nu}$ where $u=331text{ m/s}$ is the speed of sound. Putting this into the 1st equation we get

$$
Wsintheta=frac{u}{nu}Rightarrowsintheta=frac{u}{Wnu}Rightarrow theta=arcsinleft(frac{u}{Wnu}right)
$$
which is finally

$$
theta=arcsinleft(frac{331{ m/s}}{0.84text{ m}times 1300text{ Hz}}right) =17.64^circ.
$$