Is L = {  | M is a TM and L(M)={w}} turing recognizable? And its complement?
Intuitively, $L$ is not recognizable as given $\langle M, w\rangle$, to know Whether $M$ accepts $w$ we need to simulate the run of $M$ on $w$, but $M$ need not to halt on $w$, so the simulation may not halt. This obviously smells like the halting problem. Even if we somehow know that $M$ accepts $w$, how can you check whether $w$ is the only word in $L(M)$? What about the other infinite words over $M$'s alphabet? It seems impossible to go over all of these words and verify that all of them are not in $L(M)$ and then halt.
If you consider the complement of $L$, $\overline{L} = \{ \langle M, w\rangle: L(M)\neq \{w\} \}$, you get the same problamatic issue. How can you check whether $L(M)$ does not include $w$?
So intuitively, both $L$ and $\overline{L}$ are not recognizable. Formally, to show that a language $B$ is not recognizable, we show that there a non-recognizable $A$ that reduces to $B$. We consider the non-recognizable language $A = \overline{Halt_{TM}} = \{ \langle M, w \rangle: \text{M does not halt on w}\}$, then we show that there are mapping reductions, one from $A$ to $L$, and the other is from $A$ to $\overline{L}$:

$\overline{Halt_{TM}} \leq_m \overline{L}$: we need to show that there is a reduction from $\overline{Halt_{TM}}$ to $\overline{L}$. Consider the reduction that operates as follows. On input $\langle M, w\rangle$, the reduction outputs $\langle K, w\rangle$, where $K$ is a machine that operates as follows. On input $x$ for $K$, the machine $K$ simulates the run of $M$ on $w$. If $M$ halts on $w$ (if the simulation halts), then $K$ accepts $x$ only if $x = w$, and otherwise $K$ rejects $x$. If $M$ does not halt on $w$, that is the simulation does not halt, then clearly $K$ does not halt.
The reduction is correct: if $\langle M, w\rangle \in \overline{Halt_{TM}}$, then $M$ does not halt on $w$. Hence, $K$ does not halt on any input $x$, and thus $L(K) = \emptyset \neq \{w\}$. Meaning that $\langle K, w\rangle \in \overline{L}$. Conversely, if $\langle M, w\rangle\notin \overline{Halt_{TM}}$, then  $M$ halts on $w$. Hence, $K$ halts on every input $x$, but accepts only if $x = w$, and so $L(K) = \{ w\}$. Thus, $\langle K, w\rangle \notin \overline{L}$.

$\overline{Halt_{TM}} \leq_m L$: we need to show that there is a reduction from $\overline{Halt_{TM}}$ to L. Consider the reduction that operates as follows. On input $\langle M, w\rangle$, the reduction outputs $\langle K, 1\rangle$, where $K$ is a machine that operates as follows. On input $x$ for $K$, the machine $K$ checks whether $x = 1$. If $x = 1$, $K$ accepts. If $x\neq 1$, $K$
simulates the run of $M$ on $w$. If $M$ halts on $w$ (if the simulation halts), then $K$ accepts $x$. If $M$ does not halt on $w$, that is the simulation does not halt, then clearly $K$ does not halt.
The reduction is correct: if $\langle M, w\rangle \in \overline{Halt_{TM}}$, then $M$ does not halt on $w$. Hence, for every input $x\neq 1$ for $K$, $K$ does not halt, and thus $L(K) = \{1\}$. Meaning that $\langle K, 1\rangle \in L$. Conversely, if $\langle M, w\rangle\notin \overline{Halt_{TM}}$, then  $M$ halts on $w$. Hence, $K$ halts on every input $x$ (as when $x\neq 1$, the simulation of the run of $M$ on $w$ halts), and accepts. Hence, $L(K) = \Sigma^* \neq \{ 1\}$. Thus, $\langle K, 1\rangle \notin L$.


Final note: I have assumed w.l.o.g that $\overline{L}$ does not contain words that are not a description of a pair $\langle M, w \rangle$. This is okay, as a reduction (a machine) can easily check whether an input is of the form $\langle \text{ machine, word}\rangle$. If the input of the reduction is not of this form, then the reduction outputs some constant word that is in the language that we're reducing to. So for readability issues, I sticked to this assumption.