Typically the stationary distributions will be relatively easy to find given that you find the correct simultaneous equations  however in more complex cases Gauss reduction can be used to find the correct solution.
The case of Tut 5 Q4 b iii) is solved as follows, where:

3a = Level 3 this year following level 2 last year;

3b = Level 3 this year following level 4 last year.
and the corresponding transition matrix isĀ

1

2

3a

4

3b

1

0.2

0.8

0

0

0

2

0.2

0

0.8

0

0

3a

0

0.2

0

0.8

0

4

0

0

0

0.8

0.2

3b

0.2

0

0

0.8

0

hence, the simultaneous equations are as follows:
$$\pi_1 = 0.2\pi_1 + 0.2\pi_2 +0.2\pi_{3b} \\ \pi_2 = 0.8\pi_1 +0.2\pi_{3a} \\ \pi_{3a} = 0.2\pi_{3b} + 0.2\pi_{3a} \\ \pi_{3b} = 0.2\pi_4$$ and the final condition of $$\sum_{i=1}^5 \pi_i =1$$
Solving the above through either substitution, ensuring that we use our limiting condition \(\sum_{i=1}^5 \pi_i =1\), or using Gauss reduction, we obtain \((\pi_1, \pi_2,\pi_{3a},\pi_4,\pi_{3b}) = \frac{1}{441}(21,20,16,320,64)\) and hence the longrun proportion of time spent in level 3 is \(\frac{16+61}{441} = 0.18141\).