Answer
$\,\,a.\,\, B_{1}\cup B_{2}\cup B_{3}\cup B_{4}= B_{4}=\left \{ x\in \mathbb{R}\mid 0\leq x\leq 4 \right \}$
$ b.\,\, \,\,\,B_{1}\cap B_{2}\cap B_{3}\cap B_{4}= B_{1}=\left \{ x\in \mathbb{R}\mid 0\leq x\leq 1 \right \}$
$c.\,\,B_{1},B_{2},B_{3},B_{4}\,\,are\,\,not\,\,mutually\,\, disjoint$
Work Step by Step
$B_{i}=\left \{ x\in \mathbb{R}\mid 0\leq x\leq i \right \} for \,\, i =1,2,3,4 $
$B_{1}=\left \{ x\in \mathbb{R}\mid 0\leq x\leq 1 \right \}$
$B_{2}=\left \{ x\in \mathbb{R}\mid 0\leq x\leq 2 \right \}$
$B_{3}=\left \{ x\in \mathbb{R}\mid 0\leq x\leq 3 \right \} $
$B_{4}=\left \{ x\in \mathbb{R}\mid 0\leq x\leq 4 \right \} $
$we\,\, notice\,\, that\,\, B_{1}\subseteq B_{2}\subseteq B_{3}\subseteq B_{4} $
$so\,\,a.\,\, B_{1}\cup B_{2}\cup B_{3}\cup B_{4}= B_{4}=\left \{ x\in \mathbb{R}\mid 0\leq x\leq 4 \right \}$
$and\,\, b.\,\, \,\,\,B_{1}\cap B_{2}\cap B_{3}\cap B_{4}= B_{1}=\left \{ x\in \mathbb{R}\mid 0\leq x\leq 1 \right \}$
$B_{1},B_{2},B_{3},B_{4}\,\,are\,\,not\,\,mutually\,\, disjoint\,\,as\,\,for\,example\,$
$B_{2}\cap B_{4}=\left \{ x\in \mathbb{R}\mid 0\leq x\leq 2 \right \}\cap \left \{ x\in \mathbb{R}\mid 0\leq x\leq 4 \right \}=\left \{ x\in \mathbb{R}\mid 0\leq x\leq 2 \right \}\neq \varnothing $