$lub(a, b)$ and $glb(a, b)$ of non- comparable lattice elements In an algebraic lattice $(L, \vee, \wedge, 0, 1)$, the binary operations $\vee$ and $\wedge$ are commutative and associative, satisfying absorption laws. The elements $0$ and $1$ are top and bottom elements, respectively. In other words $0$ is additive identity and $1$ is multiplicative identity.
In general, elements $a,b\in L$ are comparable if $a\leq b$ and in that case, least upper bound ($lub$)of $a$ and $b$= $ a\vee b=b$ and greatest  lower bound ($glb$) of $a$ and $b$=$ a\wedge b=a$
However, if $a$ and $b$ are not comparable, then presumably $lub(a, b)$ =$ a\vee b=1$ and $glb(a, b)$=$ a\wedge b=0$.

Can we find an example of lattice in which $lub(a, b)$ =$ a\vee b\neq 1$ and $glb(a, b)$=$ a\wedge b\neq 0$ when $a\nleq b$?

 A: Your nomenclature is off: $a\wedge b$ is called the greatest lower bound of $a$ and $b$, or the “meet of $a$ and $b$”, not the infimum of $a\wedge b$; $a\vee b$ is called the least upper bound of $a$ and $b$ or the “join of $a$ and $b$”, not the supremum of $a\vee b$. For complete lattices, if $S$ is a subset, then $\bigvee S$ is the supremum of $S$, $\bigwedge S$ the infimum of $S$. You could say $a\vee b$ is “the supremum of $a$ and $b$”, but not “the supremum of $a\vee b$”.
Finally, note that “algebraic lattice” is actually a term of art: it means a complete lattice in which every element is a join of compact elements. I think what you mean is “a lattice viewed as an algebra (in the sense of universal algebra)”.
As to the rest, you are somewhat confused. If $a$ and $b$ are comparable, then $a\wedge b = \min\{a,b\}$ and $a\vee b= \max\{a,b\}$. So if, for example, $a\leq b$, then $a\vee b = b$ and $a\wedge b = a$. In particular, if you took $a=0$ and $b=1$, you would have $a\wedge b =0$ and $a\vee b = 1$; your assertion that if they are comparable then $a\wedge b\neq 0$ and $a\vee b\neq 1$ is false.
As to examples: let $L$ be the lattice of subsets of $\{1,2,3,4\}$ ordered by inclusion. Here, $A\wedge B = A\cap B$, $A\vee B = A\cup B$, $\mathbf{0}=\varnothing$, and $\mathbf{1}=\{1,2,3,4\}$.
Let $a=\{1,2\}$, $b=\{2,3\}$. These two elements are incomparable, since neither is contained in the other. Here, $a\wedge b = a\cap b = \{2\}\neq\varnothing$, and $a\vee b = a\cup b = \{1,2,3\}\neq\{1,2,3,4\}$.
In a $01$-lattice, two elements $a$ and $b$ such that $a\wedge b = 0$ and $a\vee b=1$ are said to be “complementary”.
