`[(a' + b')(a + c')(b' + c')]'`

{taking complement of the given expression}

`(a' + b')' + (a + c')' + (b' + c')'`

{from De Morgan’s law}

`(a')'(b')' + (a)'(c')' + (b')'(c')'`

{from De Morgans law}

`ab + a'c + bc`

{complement of x’ = x}

`ab + a'c`

{from Redundancy Theorem}

To know about Redundancy Theorem follow the link: https://youtu.be/3pbH9IhxwOg

If you don’t know about Redundancy Theorem then you may use the solution below

`ab + a'c + bc.1`

{since x.1 = x}

`ab + a'c + bc(a' + a)`

{since x’ + x = 1}

`ab + a'c + a'bc + abc`

{from Distributive Law}

`ab(1 + c) + a'c(1 + b)`

{taking ab common from the first and last terms and taking a’c common from the two middle terms}

`ab.1 + a'c.1`

{since 1 + x = 1}

`ab + a'c`

{since 1.x = x}

The b part of the problem is not based on Boolean Algebra so please ask it in math section.