Answer: Option A. -> 17
$$\eqalign{
& {4^{61}} + {4^{62}} + {4^{63}} + {4^{64}} \cr
& = {4^{61}}\left( {{4^0} + {4^1} + {4^2} + {4^3}} \right) \cr
& = {4^{61}} \times 85 \cr} $$
Now check with option 85 is divisible by 17.

Answer: Option B. -> 1
$${x^{\left( {b - c} \right)\left( {b + c - a} \right)}}.{x^{\left( {c - a} \right)\left( {c + a - b} \right)}}.{x^{\left( {a - b} \right)\left( {a + b - c} \right)}}$$
  $$ = {x^{\left( {b - c} \right)\left( {b + c} \right) - a\left( {b - c} \right)}}.$$    $${x^{\left( {c - a} \right)\left( {c + a} \right) - b\left( {c - a} \right)}}.$$   $${x^{\left( {a - b} \right)\left( {a + b} \right) - c\left( {a - b} \right)}}$$
$$\eqalign{
& = {x^{\left( {{b^2} - {c^2} + {c^2} - {a^2} + {a^2} - {b^2}} \right)}}.{x^{ - a\left( {b - c} \right) - b\left( {c - a} \right) - c\left( {a - b} \right)}} \cr
& = \left( {{x^0} \times {x^0}} \right) \cr
& = \left( {1 \times 1} \right) \cr
& = 1 \cr} $$

Answer: Option B. -> 1
$$\eqalign{
& {a^1} \cr
& = {c^z} \cr
& = {\left( {{b^y}} \right)^z} \cr
& = {b^{yz}} \cr
& = {\left( {{a^x}} \right)^{yz}} \cr
& = {a^{xyz}} \cr
& \Rightarrow xyz = 1 \cr} $$