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Let me prove $(1)$.

First of all, I guess that $f, \, g$ are homogeneous polynomials of the same degree $d$, otherwise $Z(f+ \varepsilon g)$ is not defined as a subvariety of $\mathbb{RP}^k$.

That said, note that the locus $S_d$ of singular hypersurfaces of degree $d$ in $\mathbb{RP}^k$ is closed in the locus of all hypersurfaces, because it is given by a finite number of polynomial equations (obtained imposing the vanishing of all $k+1$ partial derivatives).

Since, by assumption, $Z(f+\varepsilon g)$ is smooth for $\varepsilon =0$ and $S_d$ is closed, it follows that $Z(f + \varepsilon g)$ is smooth for $|\varepsilon|<t$, where $t$ is sufficiently small. A similar continuity argument shows that $Z(f + \varepsilon g)$ has still codimension $1$ for $t$ sufficiently small (and I guess that this "sufficiently small" can be made explicit in terms of grad$(f)$, also answering $(2)$).

Now, set $$\mathscr{X}=\{(x, \, t) \, | \, f(x)+tg(x)=0 \} \subset \mathbb{RP}^k \times (-t, \, t).$$

Then, projecting over the factor $(-t, \,t)$ we have a surjective smooth submersion $F \colon \mathscr{X} \to (-t, \, t)$, whose (compact) fibre $X_{\varepsilon}=F^{-1}(\varepsilon)$ is precisely $Z(f + \varepsilon g)$.

By Ehresmann Lemma, it follows that $F$ is a locally trivial fibration and so all its fibres are diffeomorphic.

Let me prove $(1)$.

First of all, I guess that $f, \, g$ are homogeneous polynomials of the same degree $d$, otherwise $Z(f+ \varepsilon g)$ is not well-defined as a subvariety of $\mathbb{RP}^k$.

That said, note that the locus $S_d$ of singular hypersurfaces of degree $d$ in $\mathbb{RP}^k$ is closed in the locus of all hypersurfaces, because it is given by a finite number of polynomial equations (obtained imposing the vanishing of the $k+1$ partial derivatives).

Since, by assumption, $Z(f+\varepsilon g)$ is smooth for $\varepsilon =0$ and $S_d$ is closed, it follows that $Z(f + \varepsilon g)$ is smooth for $|\varepsilon|<t$, where $t$ is sufficiently small. A similar continuity argument shows that $Z(f + \varepsilon g)$ has still codimension $1$ for $t$ sufficiently small (and I guess that this "sufficiently small" can be made explicit in terms of grad$(f)$, also answering $(2)$).

Now, set $$\mathscr{X}=\{(x, \, t) \, | \, f(x)+tg(x)=0 \} \subset \mathbb{RP}^k \times (-t, \, t).$$

Then, projecting over the factor $(-t, \,t)$ we have a surjective smooth submersion $F \colon \mathscr{X} \to (-t, \, t)$, whose (compact) fibre $X_{\varepsilon}=F^{-1}(\varepsilon)$ is precisely $Z(f + \varepsilon g)$.

By Ehresmann Lemma, it follows that $F$ is a locally trivial fibration and so all its fibres are diffeomorphic.

Let me prove $(1)$.

First of all, I guess that $f, \, g$ are homogeneous polynomials of the same degree $d$, otherwise $Z(f+ \varepsilon g)$ is not defined as a subvariety of $\mathbb{RP}^k$.

That said, note that the locus $S_d$ of singular hypersurfaces of degree $d$ in $\mathbb{RP}^k$ is open in the locus of all hypersurfaces, because it is given by a finite number of polynomial equations (obtained imposing the vanishing of all $k+1$ partial derivatives).

Since, by assumption, $Z(f+\varepsilon g)$ is smooth for $\varepsilon =0$ and $S_d$ is closed, it follows that $Z(f + \varepsilon g)$ is smooth for $|\varepsilon|<t$, where $t$ is sufficiently small. A similar continuity argument shows that $Z(f + \varepsilon g)$ has still codimension $1$ for $t$ sufficiently small (and I guess that this "sufficiently small" can be made explicit in terms of grad$(f)$, also answering $(2)$).

Now, set $$\mathscr{X}=\{(x, \, t) \, | \, f(x)+tg(x)=0 \} \subset \mathbb{RP}^k \times (-t, \, t).$$

Then, projecting over the factor $(-t, \,t)$ we have a surjective smooth submersion $F \colon \mathscr{X} \to (-t, \, t)$, whose (compact) fibre $X_{\varepsilon}=F^{-1}(\varepsilon)$ is precisely $Z(f + \varepsilon g)$.

By Ehresmann Lemma, it follows that $F$ is a locally trivial fibration and so all its fibres are diffeomorphic.

Let me prove $(1)$.

First of all, I guess that $f, \, g$ are homogeneous polynomials of the same degree $d$, otherwise $Z(f+ \varepsilon g)$ is not defined as a subvariety of $\mathbb{RP}^k$.

That said, note that the locus $S_d$ of singular hypersurfaces of degree $d$ in $\mathbb{RP}^k$ is closed in the locus of all hypersurfaces, because it is given by a finite number of polynomial equations (obtained imposing the vanishing of all $k+1$ partial derivatives).

Since, by assumption, $Z(f+\varepsilon g)$ is smooth for $\varepsilon =0$ and $S_d$ is closed, it follows that $Z(f + \varepsilon g)$ is smooth for $|\varepsilon|<t$, where $t$ is sufficiently small. A similar continuity argument shows that $Z(f + \varepsilon g)$ has still codimension $1$ for $t$ sufficiently small (and I guess that this "sufficiently small" can be made explicit in terms of grad$(f)$, also answering $(2)$).

Now, set $$\mathscr{X}=\{(x, \, t) \, | \, f(x)+tg(x)=0 \} \subset \mathbb{RP}^k \times (-t, \, t).$$

Then, projecting over the factor $(-t, \,t)$ we have a surjective smooth submersion $F \colon \mathscr{X} \to (-t, \, t)$, whose (compact) fibre $X_{\varepsilon}=F^{-1}(\varepsilon)$ is precisely $Z(f + \varepsilon g)$.

By Ehresmann Lemma, it follows that $F$ is a locally trivial fibration and so all its fibres are diffeomorphic.

Let me prove $(1)$.

First of all, I guess that $f, \, g$ are homogeneous polynomials of the same degree $d$, otherwise $Z(f+ \varepsilon g)$ is not defined as a subvariety of $\mathbb{RP}^k$.

That said, note that the locus $S_d$ of singular hypersurfaces of degree $d$ in $\mathbb{RP}^k$ is open in the locus of all hypersurfaces, because it is given by a finite number of polynomial equations (obtained imposing the vanishing of all $d$ partial derivatives).

Since, by assumption, $Z(f+\varepsilon g)$ is smooth for $\varepsilon =0$ and $S_d$ is closed, it follows that $Z(f + \varepsilon g)$ is smooth for $|\varepsilon|<t$, where $t$ is sufficiently small. A similar continuity argument shows that $Z(f + \varepsilon g)$ has still codimension $1$ for $t$ sufficiently small (and I guess that this ``sufficiently small" can be made explicit in terms of grad$(f)$, also answering $(2)$).

Now, set $$\mathscr{X}=\{(x, \, t) \, | \, f(x)+tg(x)=0 \} \subset \mathbb{RP}^k \times (-t, \, t).$$

Then, projecting over the factor $(-t, \,t)$ we have a surjective smooth submersion $F \colon \mathscr{X} \to (-t, \, t)$, whose (compact) fibre $X_{\varepsilon}=F^{-1}(\varepsilon)$ is precisely $Z(f + \varepsilon g)$.

By Ehresmann Lemma, it follows that $F$ is a locally trivial fibration and so all its fibres are diffeomorphic.

Let me prove $(1)$.

First of all, I guess that $f, \, g$ are homogeneous polynomials of the same degree $d$, otherwise $Z(f+ \varepsilon g)$ is not defined as a subvariety of $\mathbb{RP}^k$.

That said, note that the locus $S_d$ of singular hypersurfaces of degree $d$ in $\mathbb{RP}^k$ is open in the locus of all hypersurfaces, because it is given by a finite number of polynomial equations (obtained imposing the vanishing of all $k+1$ partial derivatives).

Since, by assumption, $Z(f+\varepsilon g)$ is smooth for $\varepsilon =0$ and $S_d$ is closed, it follows that $Z(f + \varepsilon g)$ is smooth for $|\varepsilon|<t$, where $t$ is sufficiently small. A similar continuity argument shows that $Z(f + \varepsilon g)$ has still codimension $1$ for $t$ sufficiently small (and I guess that this "sufficiently small" can be made explicit in terms of grad$(f)$, also answering $(2)$).

Now, set $$\mathscr{X}=\{(x, \, t) \, | \, f(x)+tg(x)=0 \} \subset \mathbb{RP}^k \times (-t, \, t).$$

Then, projecting over the factor $(-t, \,t)$ we have a surjective smooth submersion $F \colon \mathscr{X} \to (-t, \, t)$, whose (compact) fibre $X_{\varepsilon}=F^{-1}(\varepsilon)$ is precisely $Z(f + \varepsilon g)$.

By Ehresmann Lemma, it follows that $F$ is a locally trivial fibration and so all its fibres are diffeomorphic.