Given $N$ is a set of natural numbers and $R=\{(x,y):y=x+5 and x<4\}$:

Since $x<4$ and $ y=x+5 \Rightarrow x = \{1,2,3,4\}$ and $y = \{6,7,8,9\}$

$\Rightarrow R = \{(1,6), (2,7), (3,8), (4,9)\}$

For $R = \{(1,6), (2,7), (3,8), (4,9)\}$ to be reflexive, $(a,a) \in R\;$ for every $\; a\in\;A$

$\Rightarrow$ if $y=x, x = x+5 \Rightarrow 0 \neq 5$. Therefore $R$ is not reflexive.

We can verify this w/ a simple substitution:

If $x=1, y=1, y = x+5 \rightarrow 1 = 6$, which is not correct. Therefore $R$ is not reflexive.

For $R = \{(1,6), (2,7), (3,8), (4,9)\}$ to be symmetrical $(a_1,a_2) \in R\;\Rightarrow\; (a_2,a_1)\in R \; for \;a_1,a_2 \in A$

$\Rightarrow R \{x,y\}: y = x+5 \rightarrow y - x = 5$

$\Rightarrow R \{y,x\}: x = y+5 \rightarrow y - x = -5$

Therefore, since $(x,y) \in R,$ but $(y,x) \not \in R, \; R$ is not transitive

We can verify this w/ a simple substitution:

If $x=1, y=6, R \{x,y\}: y = x+5 \rightarrow y = x+5 \rightarrow 6 = 6$

However, $R \{y,x\}: x = y+5 \rightarrow 1 = 6+5 \rightarrow 1 = 11$ which is not correct. Hence, $R$ is not symmetric.

For $R = \{(1,6), (2,7), (3,8), (4,9)\}$ to be transitive, $(a_1,a_2) \in R$ and $(a_2,a_3) \in R \; \Rightarrow \;(a_1,a_3)\in R$ for all$\; a_1,a_2,a_3 \in A$

Let's take the first ordered pair $(1,6) \in R$. Here $x=1, y=6$.

However, since $x<4$ in $R = \{x,y\}$, there can exisit no ordered pair for $(y,z)$ where $y$ can be equal to 6.

Therefore $R$ is not transitive.