$1.3,4$ ** Find all real numbers $a$ such that for any real number $x$, the value of the function $f(x)=x^{2}-2 x-|x-1-a|-|x-2|+4$ is a non-negative real number.
From the conditions, we have
\[
\left\{
\begin{array}{l}
f(0) = -|1 + a| + 2 \geqslant 0, \\
f(1) = -|a| + 2 \geqslant 0,
\end{array}
\right.
\]
solving this, we get $-2 \leqslant a \leqslant 1$.
Next, we prove that when $-2 \leqslant a \leqslant 1$, for any $x \in \mathbf{R}$, we have $f(x) \geqslant 0$.
In fact, let $t = x - 1$, then
\[
f(x) = t^2 - |t - a| - |t - 1| + 3.
\]

Let $g(t) = t^2 + 3 - |t - a| - |t - 1|$.
When $t \leqslant a$, $g(t) = t^2 + 3 - (a - t) - (1 - t) = t^2 + 2t + 2 - a = (t + 1)^2 + (1 - a) \geqslant 0$;

When $a \leqslant t \leqslant 1$, $g(t) = t^2 + 3 - (t - a) - (1 - t) = t^2 + 2 + a$, combining $a \geqslant -2$ we know $g(t) \geqslant 0$;

When $t \geqslant 1$, $g(t) = t^2 + 3 - (t - a) - (t - 1) = t^2 - 2t + 4 + a = (t - 1)^2 + 3 + a \geqslant 3 + a \geqslant 1$.
Therefore, when $-2 \leqslant a \leqslant 1$, we always have $f(x) \geqslant 0$.
The set of $a$ that satisfies the conditions is $\{a \mid -2 \leqslant a \leqslant 1\}$.