Answer:

[tex]\qquad \:\boxed{\begin{aligned}& \qquad \:\sf \: (1) \: \: 5 \qquad \: \\ \\& \qquad \:\sf \: (3) \: \: \dfrac{4}{9 {a}^{2} {x}^{2} } \end{aligned}} \qquad \\ \\ [/tex]

Step-by-step explanation:

[tex]\large\underline{\sf{Solution-1}}[/tex]

Given logarithmic equation is

[tex]\sf \: log_{2}(4) + log_{2}(3x + 1) = 6 \\ \\ [/tex]

[tex]\sf \: log_{2}[ 4(3x + 1)] = 6 \\ \\ [/tex]

[tex]\qquad\boxed{ \sf{ \: \because \: log_{a}(m) + log_{a}(n) = log_{a}(mn) \: }} \\ \\ [/tex]

[tex]\sf \: log_{2}[ 12x + 4] = 6 \\ \\ [/tex]

[tex]\sf \: 12 x + 4 = {2}^{6} \\ \\ [/tex]

[tex]\qquad\boxed{ \sf{ \: \because \: log_{a}(b) = c \: \: \sf\implies \: b = {a}^{c} \: }} \\ \\ [/tex]

[tex]\sf \: 12 x + 4 = 64 \\ \\ [/tex]

[tex]\sf \: 12 x = 64 - 4 \\ \\ [/tex]

[tex]\sf \: 12 x = 60 \\ \\ [/tex]

[tex]\sf\implies \bf \: x = 5 \\ \\ [/tex]

[tex]\large\underline{\sf{Solution-2}}[/tex]

Given that,

[tex]\sf \:\dfrac{logx}{b - c} = \dfrac{logy}{c - a} = \dfrac{logz}{a - b} \\ \\ [/tex]

Let assume that

[tex]\sf \:\dfrac{logx}{b - c} = \dfrac{logy}{c - a} = \dfrac{logz}{a - b} = k\\ \\ [/tex]

[tex]\sf \: logx = k(b - c) - - - (1) \\ \\ [/tex]

[tex]\sf \: logy = k(c - a) - - - (2) \\ \\ [/tex]

[tex]\sf \: logz = k(a - b) - - - (3) \\ \\ [/tex]

On multiply equation (1) by b + c, (2) by c + a and (3) by a + b, we get

[tex]\sf \: (b + c)logx = k(b + c)(b - c) \: \sf\implies log {x}^{b + c} = k( {b}^{2} - {c}^{2}) - - (4) \\ \\ [/tex]

[tex]\sf \: (c + a)logy = k(c + a)(c - a) \: \sf\implies log {y}^{c + a} = k( {c}^{2} - {a}^{2}) - - (5) \\ \\ [/tex]

[tex]\sf \: (a + b)logz = k(a + b)(a - b) \: \sf\implies log {z}^{a + b} = k( {a}^{2} - {b}^{2}) - - (6) \\ \\ [/tex]

On adding equation (4), (5) and (6), we get

[tex]\sf \: log {x}^{b + c} + log {y}^{c + a} + log {z}^{a + b} = k( {b}^{2} - {c}^{2} + {c}^{2} - {a}^{2} + {a}^{2} - {b}^{2}) \\ \\ [/tex]

[tex]\sf \: log\left( {x}^{b + c} {y}^{c + a} {z}^{a + b}\right) = k( 0) \\ \\ [/tex]

[tex]\qquad\boxed{ \sf{ \: \because \: log_{a}(m) + log_{a}(n) = log_{a}(mn) \: }} \\ \\ [/tex]

[tex]\sf \: log\left( {x}^{b + c} {y}^{c + a} {z}^{a + b}\right) = 0 \\ \\ [/tex]

[tex]\sf \: log\left( {x}^{b + c} {y}^{c + a} {z}^{a + b}\right) = log1 \\ \\ [/tex]

[tex]\sf\implies \bf \: {x}^{b + c} {y}^{c + a} {z}^{a + b} = 1 \\ \\ \\ [/tex]

[tex]\large\underline{\sf{Solution- 3}}[/tex]

Given expression is

[tex]\sf \: {\left(\dfrac{27 {x}^{3} }{8 {a}^{ - 3} } \right)}^{ - \frac{2}{3} } \\ \\ [/tex]

[tex]\sf \: = \: {\left(\dfrac{3.3.3 .{x}^{3} {a}^{3} }{2.2.2 } \right)}^{ - \frac{2}{3} } \\ \\ [/tex]

[tex]\sf \: = \: {\left(\dfrac{ {3}^{3} .{x}^{3} .{a}^{3} }{ {2}^{3} } \right)}^{ - \frac{2}{3} } \\ \\ [/tex]

[tex]\sf \: = \: {\left(\dfrac{ 3ax}{ 2 } \right)}^{ - \frac{2}{3} \times 3} \\ \\ [/tex]

[tex]\sf \: = \: {\left(\dfrac{ 3ax}{ 2 } \right)}^{ - 2} \\ \\ [/tex]

[tex]\sf \: = \: {\left(\dfrac{2}{ 3ax } \right)}^{ 2} \\ \\ [/tex]

[tex]\sf \: = \: \dfrac{4}{9 {a}^{2} {x}^{2} } \\ \\ [/tex]

Hence,

[tex]\sf\implies \bf \: {\left(\dfrac{27 {x}^{3} }{8 {a}^{ - 3} } \right)}^{ - \frac{2}{3} } = \dfrac{4}{9 {a}^{2} {x}^{2} } \\ \\ [/tex]