For the following, I take 3 to have one significant figure, 67 to have 2 significant figures, and 100 to have one significant figure. If you want to avoid confusion with exact integers (such as a stoichiometric coefficient), use scientific notation for numbers lacking a decimal point, as in $\pu{0.3e1}, \pu{6.7e1} \text{ and }\pu{0.1e3}$.
Pencil and paper
The step-by-step approach is to consider each step separately (observing the correct order of priority), keeping track of significant figures of the intermediate results. Then, calculate the number without any rounding of the intermediate results, and record it with the appropriate number
(5.01 + 4.1) / (1.00 x 1.00)
- The sum will be significant to a tenth, so 2 significant figures.
- The product will have 3 significant figures because that is the minimum of the factors
- For the division, the numerator has 2 significant figures and the denominator has 3, so the result has 2 significant figures.
- The arithmetic result without any rounding is 9.11, so you write down 9.1.
(2.00 x 30) + 5.01 + 2.0
- The product has one significant figures, least significant figure is the tens.
- Adding 5.01 will have least significant figure as the tens.
- Adding 2.0 will have least significant figure as the tens.
- The arithmetic result (unrounded) is 67.01, so you write down 70 or $\pu{0.7e2}$.
(5.01 + 2.0 - 3) / (60.01 x 3.0)
- The sum of three numbers will be significant to the ones, giving one significant figure.
- The product will have 2 significant figures because 3.0 does too.
- The overall answer will have 1 significant figure because the numerator does too.
- The arithmetic result (unrounded) is 0.02227406543..., so you write down 0.02.
Using a tool
I wrote a calculator in a browser that knows about significant figures (and units), described in J. Chem. Educ.2015, 9211, 1953-1955. For the first example, it gives the following output (input is:"(5.01 + 4.1) / (1.00 * 1.00)"):
$$\mathrm{result} = \dfrac{5.01 + 4.1}{1.00 \cdot 1.00}$$
$$\ \ \ =\dfrac{9.11}{1.000}$$
$$=9.1\ \ $$
The intermediate results are written with one extra digit ("guard digit") so that no information is lost (in the pencil and paper section above, I kept all digits until rounding at the very end). If you write down and then reuse intermediate results, you should follow that practice (unless your teacher tells you otherwise). The calculator is available at http://pqcalc.pythonanywhere.com/. Behind the scenes, it uses a slightly more sophisticated error propagation model, so there are times when it shows a digit more than you would expect. Also, different from most textbooks, the calculator interprets numbers without decimal points as exact integers, and numbers with a trailing decimal point (67. or 100.) as significant to the ones position.