To evaluate the given expression, we use PEMDAS.

Notice that $11^0=1.$ So, we obtain:

\begin{align*} 8^2 \div 16 + 11^0 &=\\[3pt] 64 \div 16+1 &=\\[3pt] (64 \div 16)+1 &=\\[3pt] 4 + 1 &=\\[3pt] 5 \end{align*}

To evaluate the given expression, we use PEMDAS.

Notice that $2^0=1.$ So, we obtain:

\begin{align*} -(3)^2+(-5) \times 2^0 &=\\[3pt] -9 + (-5) \times 1 &=\\[3pt] -9 + (-5) &=\\[3pt] -9 - 5 &=\\[3pt] -14 \end{align*}

To evaluate the given expression, we use PEMDAS.

Notice that $\left(-\dfrac{4}{3}\right)^{\!0}=1.$ So, we obtain:

\begin{align*} \left(-\dfrac{4}{3}\right)^{\!0} \div (1+3^2) - 3.5 &=\\[5pt] \left(-\dfrac{4}{3}\right)^{\!0} \div (1+9) - 3.5 &=\\[5pt] \left(-\dfrac{4}{3}\right)^{\!0} \div 10 - 3.5 &=\\[5pt] 1 \div 10 - 3.5 &=\\[5pt] 0.1 - 3.5 &=\\[5pt] -3.4 \end{align*}

To evaluate the given expression, we use PEMDAS.

Notice that $5^{-2}=\dfrac{1}{5^2}.$ So, we obtain:

\begin{align*} \require{cancel} 5^{-2} \div \dfrac{1}{5} &=\\[5pt] \dfrac{1}{5^2} \div \dfrac{1}{5} &=\\[5pt] \dfrac{1}{25} \div \dfrac{1}{5} &=\\[5pt] \dfrac{1}{25} \times \dfrac{5}{1} &=\\[5pt] \dfrac{1 \times 5}{25 \times 1} &=\\[5pt] \dfrac{1 \times \cancel{5}}{5 \times \cancel{5}} &=\\[5pt] \dfrac{1}{5} \end{align*}

To evaluate the given expression, we use PEMDAS.

Notice that $2^{-2}=\dfrac{1}{2^2}.$ So, we obtain:

\begin{align*} \require{cancel} 2^{-2} \div \left(\dfrac{1}{4}-1\right) &=\\[5pt] \dfrac{1}{2^2} \div \left(\dfrac{1}{4}-1\right) &=\\[5pt] \dfrac{1}{4} \div \left(\dfrac{1}{4}-1\right) &=\\[5pt] \dfrac{1}{4} \div \left(-\dfrac{3}{4}\right) &=\\[5pt] \dfrac{1}{4} \times \left(-\dfrac{4}{3}\right) &=\\[5pt] -\dfrac{1 \times \cancel{4}}{\cancel{4} \times 3} &=\\[5pt] -\dfrac{1}{3} \end{align*}

To evaluate the given expression, we use PEMDAS.

First, we compute the expression inside the parentheses:

\[ \dfrac{5}{8} \times 3 - (8-10)^{-3} = \dfrac{5}{8} \times 3 - (-2)^{-3} \]

Next, notice that $(-2)^{-3}=\dfrac{1}{(-2)^3} = -\dfrac{1}{8}.$ So, we obtain:

\begin{align*} \require{cancel} \dfrac{5}{8} \times 3 - (-2)^{-3} &=\\[5pt] \dfrac{5}{8} \times 3 - \left(-\dfrac{1}{8}\right) &=\\[5pt] \dfrac{5}{8} \times \dfrac{3}{1} + \dfrac{1}{8} &=\\[5pt] \dfrac{5 \times 3}{8 \times 1} + \dfrac{1}{8} &=\\[5pt] \dfrac{15}{8} + \dfrac{1}{8} &=\\[5pt] \dfrac{15+1}{8} &=\\[5pt] \dfrac{16}{8} &=\\[5pt] 2 \end{align*}

To evaluate the given expression, we use PEMDAS.

First, we compute the expression inside the parentheses:

\[ \left(\dfrac{5}{8}\right)^0 \times (3.5-1.5)^{-3} = \left(\dfrac{5}{8}\right)^0 \times 2^{-3} \]

Next, notice that $\left(\dfrac{5}{8}\right)^0 = 1.$ So, we obtain:

\begin{align*} \left(\dfrac{5}{8}\right)^0 \times 2^{-3} &= \\[5pt] 1 \times 2^{-3} &= \\[5pt] 2^{-3} \end{align*}

Finally, \[ 2^{-3} = \dfrac{1}{2^3} = \dfrac{1}{8}. \]

To evaluate the given expression, we use PEMDAS.

Any number raised to the power of $0$ equals $1.$ So, we have

\[ \left( \dfrac{1}{101}-0.01 \right) ^0 = 1. \]

Next, we evaluate the negative exponent and get

\[ \left(\dfrac{1}{3}\right)^{-2} =\left(\dfrac{3}{1}\right)^{2} = 3^2 = 9. \]

Substituting all these values into the expression, we obtain:

\begin{align*} 18 \div \left(\dfrac{1}{3}\right)^{-2} + \left( \dfrac{1}{101}-0.01 \right) ^0 &=\\[5pt] 18 \div 9 + 1 &=\\[5pt] 2 + 1 &=\\[5pt] 3 \end{align*}

To evaluate the given expression, we use PEMDAS.

First, notice that $ \dfrac{8}{7} - \dfrac{1}{16} $ is not zero. Any non-zero number raised to the power of $0$ equals $1.$ So, we have

\[ \left( \dfrac{8}{7} - \dfrac{1}{16} \right) ^0 = 1. \]

Next, we evaluate the negative exponent and get

\[ \left(\dfrac{1}{2}\right)^{-3} =\left(\dfrac{2}{1}\right)^{3} = 2^3 = 8. \]

Substituting all these values into the expression, we obtain:

\begin{align*} \left( \dfrac{8}{7} - \dfrac{1}{16} \right) ^0- 11 \times \left(\dfrac{1}{2}\right)^{-3}&=\\[5pt] 1-11 \times 8&=\\[5pt] 1-88 &=\\[5pt] -87 \end{align*}