Given that when P(x) is divided by (x+2), the remainder is 3, we have P(-2) = 3.

Let the polynomial be $P(x)$.
According to the Remainder Theorem, if a polynomial $P(x)$ is divided by $(x-c)$, the remainder is $P(c)$.
Given that when $P(x)$ is divided by $(x+2)$, the remainder is 3, we have $P(-2) = 3$.
Given that when $P(x)$ is divided by $(x-1)$, the remainder is $-1$, we have $P(1) = -1$.

When $P(x)$ is divided by $(x+2)(x-1)$, the divisor is a quadratic polynomial. Therefore, the remainder must be a linear polynomial of the form $Ax + B$.
We can write $P(x)$ as:

where $Q(x)$ is the quotient and $Ax + B$ is the remainder.

Substitute $x = -2$ into the equation for $P(x)$:

Substitute $x = 1$ into the equation for $P(x)$:

We have the system:

1. $-2A + B = 3$
2. $A + B = -1$ Subtract equation (2) from equation (1):

Substitute the value of $A$ into equation (2):

The remainder is $Ax + B$. Substitute the values of $A$ and $B$:

The remainder when the polynomial is divided by $(x+2)(x-1)$ is $\boxed{-\frac{4}{3}x + \frac{1}{3}}$.