The context is an integer programming problem on choosing projects to maximise profit.
I think (b) i. changes if 'either project 1 or project 3' means 'project 1 xor project 3' rather than 'project 1 or project 3'
What I tried:
Let $x_i =$
$1$, if project i is selected and $0$, otherwise.
and
$c_{ij}$ be the expenditure in project i in year j
Obj func: Max
$$z = r \cdot x - \sum_{i} x_ic_{i1} - \sum_{i} x_ic_{i2} - \sum_{i} x_ic_{i3}$$
where
$r_i$ is the return of project i
$$r' = [10, 40, 20, 15, 30]$$
$$x' = [x_1, x_2, x_3,x_4, x_5]$$
s.t.
$$\sum_{i} x_{i}c_{i1} \le 25$$
$$\sum_{i} x_{i}c_{i2} \le 25$$
$$\sum_{i} x_{i}c_{i3} \le 25$$
bii
$$x_2 + x_4 \le 1$$
bi
if either-or means or (inclusive or):
$$x_4 \ge x_1, x_4 \ge x_3$$
if either-or means xor (exclusive or):
$$x_4 \ge x_1 + x_3 - y$$
where $y=2$ if $x_1 = x_3 = 1$ and $0$ otherwise.
Is that right?
From Chapter 3 here.
For inclusive or:
$$x_4 \ge x_1 \ \text{and} \ x_4 \ge x_3$$
For exclusive or:
$$x_3 + x_4 \ge x_1 \ \text{and} \ x_1 + x_4 \ge x_3$$
Proofs based on this:
For inclusive or:
$$(x_1 \bigvee x_3) \rightarrow x_4$$
$$\iff \neg (x_1 \bigvee x_3) \vee x_4$$
$$\iff (\neg x_1 \bigwedge \neg x_3) \vee x_4$$
$$\iff (\neg x_1 \vee x_4) \bigwedge (\neg x_3 \vee x_4) $$
$$\iff (1-x_1) + x_4 \ge 1 \ \text{and} \ (1-x_3) + x_4 \ge 1$$
$$\iff x_4 \ge x_1 \ \text{and} \ x_4 \ge x_3$$
For exclusive or:
$$(x_1 \bigvee x_3) \bigwedge \neg (x_1 \bigwedge x_3) \rightarrow x_4$$
$$\iff (x_1 \bigwedge \neg x_3) \bigvee (x_3 \bigwedge \neg x_1 ) \rightarrow x_4$$
$$\vdots$$
$$(\neg x_1 \bigvee x_3 \bigvee x_4) \bigwedge (\neg x_3 \bigvee x_1 \bigvee x_4)$$
$$(1 - x_1) + x_3 + x_4 \ge 1 \ \text{and} \ (1 - x_3) + x_1 + x_4 \ge 1$$
$$x_3 + x_4 \ge x_1 \ \text{and} \ x_1 + x_4 \ge x_3$$