To find the coefficient of \(x^3\) in the expansion of \((a+x)^5 + (1-2x)^6\), we need to consider the individual expansions.

For \((a+x)^5\), the term containing \(x^3\) is given by the binomial coefficient:

\(\binom{5}{3} a^2 x^3 = 10a^2 x^3\)

For \((1-2x)^6\), the term containing \(x^3\) is:

\(\binom{6}{3} (1)^3 (-2x)^3 = 20(-8)x^3 = -160x^3\)

The total coefficient of \(x^3\) in the expansion is:

\(10a^2 - 160 = 90\)

Solving for \(a\):

\(10a^2 = 250\)

\(a^2 = 25\)

\(a = 5\)

The coefficient of \(x^4\) in the expansion of \((x + a)^6\) is given by \(\binom{6}{4} a^2 = 15a^2\). Thus, \(p = 15a^2\).

The coefficient of \(x^2\) in the expansion of \((ax + 3)^4\) is given by \(\binom{4}{2} (ax)^2 3^2 = 54a^2\). Thus, \(q = 54a^2\).

We know that \(p + q = 276\), so:

\(15a^2 + 54a^2 = 276\)

\(69a^2 = 276\)

\(a^2 = \frac{276}{69} = 4\)

\(a = \pm 2\)

Consider the binomial expansion of \(\left( \frac{x}{a} + \frac{a}{x^2} \right)^7\).

The general term is given by:

\(T_k = \binom{7}{k} \left( \frac{x}{a} \right)^{7-k} \left( \frac{a}{x^2} \right)^k\)

Simplifying, we have:

\(T_k = \binom{7}{k} \frac{x^{7-k}}{a^{7-k}} \frac{a^k}{x^{2k}} = \binom{7}{k} \frac{a^{k-(7-k)}}{x^{2k-(7-k)}} = \binom{7}{k} \frac{a^{2k-7}}{x^{k+7-2k}}\)

For the coefficient of \(x^4\), set \(k+7-2k = 4\), giving \(k = 3\).

The coefficient of \(x^4\) is:

\(\binom{7}{3} \frac{a^{6-7}}{x^4} = \frac{35}{a}\)

For the coefficient of \(x\), set \(k+7-2k = 1\), giving \(k = 6\).

The coefficient of \(x\) is:

\(\binom{7}{6} \frac{a^{12-7}}{x} = 7a^5\)

Given:

\(\frac{\frac{35}{a}}{7a^5} = 3\)

Simplifying:

\(\frac{5}{a^6} = 3\)

\(a^6 = \frac{5}{3}\)

\(a^2 = \frac{1}{9}\)

\(a = \pm \frac{1}{3}\)

The coefficient of \(x^2\) in \(\left( 1 + \frac{2}{p} x \right)^5\) is given by:

\(\binom{5}{2} \left( \frac{2}{p} \right)^2 = 10 \times \frac{4}{p^2} = \frac{40}{p^2}\)

The coefficient of \(x^2\) in \((1 + px)^6\) is given by:

\(\binom{6}{2} (p)^2 = 15p^2\)

Setting the sum of these coefficients equal to 70:

\(\frac{40}{p^2} + 15p^2 = 70\)

Multiply through by \(p^2\) to clear the fraction:

\(40 + 15p^4 = 70p^2\)

Rearrange to form a quadratic in \(p^2\):

\(15p^4 - 70p^2 + 40 = 0\)

Let \(q = p^2\), then:

\(15q^2 - 70q + 40 = 0\)

Using the quadratic formula \(q = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

\(q = \frac{70 \pm \sqrt{70^2 - 4 \times 15 \times 40}}{30}\)

\(q = \frac{70 \pm \sqrt{4900 - 2400}}{30}\)

\(q = \frac{70 \pm \sqrt{2500}}{30}\)

\(q = \frac{70 \pm 50}{30}\)

This gives \(q = 4\) or \(q = \frac{2}{3}\).

Thus, \(p^2 = 4\) or \(p^2 = \frac{2}{3}\).

Therefore, \(p = \pm 2\) or \(p = \pm \frac{\sqrt{2}}{3}\).

To find the coefficient of \(x^3\) in the expansion of \(\left(p + \frac{1}{p}x\right)^4\), we use the binomial theorem:

\(\binom{4}{3} \cdot p^{4-3} \cdot \left(\frac{1}{p}\right)^3 \cdot x^3 = 4 \cdot p \cdot \frac{1}{p^3} \cdot x^3 = \frac{4}{p^2}x^3\).

The coefficient of \(x^3\) is \(\frac{4}{p^2}\).

We are given that this coefficient is 144:

\(\frac{4}{p^2} = 144\).

Solving for \(p^2\), we get:

\(p^2 = \frac{4}{144} = \frac{1}{36}\).

Taking the square root of both sides, we find:

\(p = \pm \frac{1}{6}\).