Replacing $u$ with $3$ in the expression, we get

\begin{align*} 6 + 2u &= \\[5pt] 6 + 2(3) &= \\[5pt] 6 + 6 &= \\[5pt] 12. \end{align*}

Replacing $y$ with $2$ in the expression, we get

\begin{align*} 4y - 13 &= \\[5pt] 4(2) - 13 &= \\[5pt] 8 - 13 &= \\[5pt] -5. \end{align*}

Replacing $k$ with $4$ in the expression, we get

\begin{align*} -3k + 20 &= \\[5pt] -3(4) + 20 &= \\[5pt] -12 + 20 &= \\[5pt] 8. \end{align*}

Replacing $u$ with $2$ in the expression, we get

\begin{align*} 4 + \dfrac{u}3 &=\\[5pt] 4 +\dfrac23 &= \\[5pt] \dfrac{4\cdot3+2}{3} &=\\[5pt] \dfrac{14}3. \end{align*}

Replacing $u$ with $5$ in the expression, we get

\[ \begin{align} \dfrac{u}{2}-2 &= \\[5pt] \dfrac{5}{2}-2 &= \\[5pt] \dfrac{5-2\cdot 2}{2} &= \\[5pt] \dfrac{1}{2} . \end{align} \]

Replacing $m$ with $6$ in the expression, we get

\[ \begin{align} 3-\dfrac{m}{3} &= \\[5pt] 3-\dfrac{6}{3} &= \\[5pt] 3-2 &= \\[5pt] 1 . \end{align} \]

We replace $n$ with $-1$ in the expression and evaluate the parentheses first: \[ \begin{align} 5(n - 1) &=\\[3pt] 5(-1-1) &=\\[3pt] 5(-2) &=\\[3pt] -10 \end{align} \]

We replace $h$ with $2$ in the expression and evaluate the parentheses first:

\[ \begin{align} 7+(3+2h) &= \\[3pt] 7+(3+2\cdot 2) &= \\[3pt] 7+(3+4) &= \\[3pt] 7+ 7 &= \\[3pt] 14 \end{align} \]

We replace $x$ with $-1$ in the expression and evaluate the parentheses first:

\[ \begin{align} 1+(2-3x) &= \\[3pt] 1+(2-3\cdot (-1)) &= \\[3pt] 1+(2-(-3)) &= \\[3pt] 1+ (2+3) &= \\[3pt] 1+5 &= \\[3pt] 6 \end{align} \]

Replacing $t$ with $-2$ and $l$ with $3,$ we get

\begin{align*} 2t - l &= \\[5pt] 2(-2) - (3)&= \\[5pt] -4-3 &=\\[5pt] -7. \end{align*}

In this case, we replace $x$ with $0$ and $y$ with $0$:

\begin{align*} 5x -y + 3 &=\\[5pt] 5\cdot 0 - (0) + 3&=\\[5pt] 0-0+3&=\\[5pt] 3 \end{align*}