Higher direct image of a locally constant sheaf

Question

I am trying (struggling) with understanding the local system structure of the higher direct image of a locally constant sheaf.

Say I have a locally trivial fibration $f: X\longrightarrow B$, with fiber $F$, and a locally constant sheaf, say $\mathbb{Z}$ over $X$. I would like to understand the sheaf $R^1f_*\mathbb{Z}$.

The sheaf $R^1f_*\mathbb{Z}$ is locally constant, and should have stalk $H^1(F,\mathbb{Z})$ at every point. What I am struggling with understanding, is the local system structure on this sheaf.

For example, say I look at the points $b_1,b_2 \in B$, and assume $\gamma\in \pi_1(B,p_1,p_2)$ is a (homotopy class of) a path from $b_1$ to $b_2$.

Then $\gamma$ induces a map $$\gamma*: R^1f_*\mathbb{Z}_{b_1}\longrightarrow R^1f_*\mathbb{Z}_{b_2}.$$

1. I know that the two stalks are isomorphic to $H^1(F,\mathbb{Z})$, so that $\gamma^*$ can be identified as an endomorphism of $H^1(F,\mathbb{Z})$. Should the endomorphism induced by $\gamma^*$ be trivial or can it be non-trivial in general?

2. Assume $B$ is locally simply connected, and assume that we set $b_2 = P_1$, and choose a sequence of points on our path $\gamma$ from $b_1$ to $b_2$ that approach ever so closely to $b_1$. The truncations of the path $\gamma$ would result in a sequence of paths $\gamma_i$ connecting $b_1$ with $P_i$, and in the limit, I would imagine that through some notion of continuity, the endomorphism $\gamma_i^*$ should become closer and closer to the identity. Is that correct?

2.1 Under the previous assumptions, should a continuity argument imply that for the case of $F = S^1$, for example, where $H^1(S^1,\mathbb{Z}) = \mathbb{Z}$, that the monodromy action $\pi_1(B,b)$ on $R^1f_*\mathbb{Z}_b$ is trivial?

3. Under the previous assumptions, I understand that the monodromy action is determined by what happens to the fundamental class of $H^1(S^1,\mathbb{Z})$, which, in turn, is determined by what does the lifted map on the fiber do to its orientation. This means that the monodromy could only act as multiplication by plus or minus one. Say I replace the locally constant sheaf $\mathbb{Z}$ on $X$ by $\mathbb{R}$. Is the monodromy action still only multiplication by plus or minus one? I think that the monodromy action on $H^1(S^1,\mathbb{R})$ factors through its action on $H^1(S^1,\mathbb{Z})$ via the identification $H^1(S^1,\mathbb{Z})\otimes \mathbb{R}$. Is this always the case? Does the monodromy action on rational/real/complex cohomology factor through its action on integral cohomology?

Thanks in advance!