Пожалуйста,помогите. Алгебра

1)
$\left \{ {{3(x-1)-2(1+x)\ \textless \ 1} \atop {3x-4\ \textgreater \ 0}} \right.$
$\left \{ {{3x-3-2-2x\ \textless \ 1} \atop {3x\ \textgreater \ 4}} \right.$
$\left \{ {{x\ \textless \ 6} \atop {x\ \textgreater \ \frac{4}{3} }} \right.$
x∈(4/3;6)
2)
(√6+√3)√12-2√6*√3=√72+√36-2√18=6√2+6-6√2=6
3)а) $6^{-2} + ( \frac{9}{4} ) ^{-1} = \frac{1}{6 ^{2} } + \frac{1}{( \frac{9}{4}) } = \frac{1}{36} + \frac{4}{9} = \frac{1}{36} + \frac{16}{36} = \frac{15}{36} = \frac{5}{12}$
б) $(2 ^{-3} ) ^{2} = 2^{-6} = \frac{1}{2 ^{6} } = \frac{1}{y} = \frac{1}{64}$
в) $\frac{7 ^{-8} * 7 ^{-9} }{7 ^{-15} } = \frac{7 ^{-17} }{ 7 ^{-15} } =7 ^{-2} = \frac{1}{7 ^{2} } = \frac{1}{49}$
4) $( \frac{3}{9 - x^{2} } + \frac{1}{x-3} ) : \frac{x}{ x^{2} - 6x +9} =( \frac{3}{(3-x)(3+x)} - \frac{3+x}{(3-x)(3+x)} ) : \frac{x}{(x-3) ^{2} } =$ $\frac{x}{(3+x)(3-x)} * \frac{(3-x) ^{2} }{x} = \frac{3-x}{3+x}$
6) $- \frac{x-8}{4} +1 \geq 0; \frac{x-8}{4} \leq 1; x-8 \leq 4; x \leq 12$
x∈(-∞;12]